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CBSE Questions for Class 12 Medical Physics Electric Charges And Fields Quiz 7 - MCQExams.com
CBSE
Class 12 Medical Physics
Electric Charges And Fields
Quiz 7
Consider the electric field due to a non-conducting infinite plane having a uniform
charge density.
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The electric field varies with distance from the plane in accordance with the inverse square law.
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The electric field is independent of the distance from the plane.
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The electric field is directly proportional to square of the distance from the plane.
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The electric field is varies inversely with the distance from the plane.
Explanation
Electric field due to a non-conducting infinite plane having uniform charge density $$(\sigma)$$ is given by $$E=\dfrac{\sigma}{2\varepsilon _{0}}$$
We can see $$E$$ is independent of distance from the plane.
Calculate the solid angle subtended by an octant of a sphere at the centre of the sphere.
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$$\dfrac{\pi}{4}$$
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$$\dfrac{\pi}{3}$$
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$$\dfrac{\pi}{2}$$
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$${\pi}$$
Explanation
An octant of sphere is each part of sphere divided by a 3-axis cartesian system plane.
Its surface area is $$S=\dfrac{1}{8}\times 4\pi r^2=\dfrac{\pi r^2}{2}$$
Solid angle is given by $$\Omega=\dfrac{S}{r^2}$$
$$=\dfrac{\pi}{2}$$
Answer-(C)
A charge $$q$$ is placed at the centre of the open end of cylindrical vessel. The flux of electric field through the surface of the vessel is
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$$0$$
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$$\dfrac{q}{\epsilon_0}$$
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$$\dfrac{q}{2\epsilon_0}$$
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$$\dfrac{2q}{\epsilon_0}$$
Explanation
Given that, A charge q is placed at the center of open end Q a cylindrical vessel,we have to find the flux through the surface of the vessel.
so, when charge Q is placed at the center of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface.
so, flux through the surface of vessel is $$\cfrac { q }{ { 2\varepsilon }_{ 0 } } $$
Hence the answer is option (c).
If the electric field in some region of space is zero,
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it implies that there is no electric charge in that region.
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it implies that there is electric charge in that region.
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it does not imply anything concrete about the charges in the region.
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it means there is discharging in the region
Explanation
If electric field in some region of space is zero, it does not imply that there is no charge in that region. From gauss law in its divergence form, $$E=\dfrac{\rho}{\varepsilon _{0}}$$
If $$E=0\: \Rightarrow \:$$ charge density $$\rho$$ is $$0$$.
It implies there are equal number of opposite charges present. So, nothing concrete can be said about charges present in that region.
A short electric dipole (which consists of two point charges $$+q \,\, and -q$$) is placed at the centre O and inside a large cube (ABCDEFGH) of length L, as shown in figure. The electric flux, emanating through the cube is:
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$$q/4 \pi \varepsilon _0L$$
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Zero
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$$q/2 \pi \varepsilon _0L$$
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$$q/3 \pi \varepsilon _0L$$
Explanation
Total charge enclosed inside the cube $$= 0$$. SO, the electric flux$$=\dfrac{q}{\epsilon_0}=0$$
Find the net flux through the cylinder.
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$$-0.125Nm^2/C$$
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$$-0.25Nm^2/C$$
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$$0.25Nm^2/C$$
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$$0.125Nm^2/C$$
Explanation
Flux through the surface $$\phi=\int \overrightarrow{E}.\overrightarrow{dS}$$
Electric field $$\overrightarrow{E}$$ is constant.
So, flux $$=E\times Area$$
$$=50\times 25\times 10^{-4}$$
$$=1250\times 10^{-4}$$
$$=0.125\: NC^{-1}m^{2}$$
What is the total electric flux leaving the surface of the sphere?
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$$1.63 \times 10^2 Nm^2/C$$
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$$1.63 \times 10^4 Nm^2/C$$
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$$1.63 \times 10^6 Nm^2/C$$
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$$1.63 \times 10^8 Nm^2/C$$
Explanation
Charge,$$q$$=surface charge density $$(\sigma) \times $$ area $$(A)$$
$$\implies q=\sigma\times 4\pi r^2$$
$$\implies q=\sigma\times 4\pi \left(\dfrac{d}{2}\right)^2$$ where $$d=diameter$$
$$\implies q=80\times 10^{-6}\times 4\pi\times \left(\dfrac{2.4}{2}\right)^2$$
$$\implies q=1.45\times 10^{-3}C=1.45mC$$
Hence, flux$$=\dfrac{q}{\epsilon_o}=\dfrac{1.45\times 10^{-3}}{8.85\times 10^{-12}}$$
$$\implies \phi=1.63\times 10^8 Nm^2/C$$
Answer-(D)
Relative permittivity of water is $$81$$. If $$\epsilon_{w}$$ and $$\epsilon_{0}$$ are permittivities of water and vacuum respectively, then :
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$$\epsilon_{0} = 9\epsilon_{w}$$
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$$\epsilon_{0} = 81\epsilon_{w}$$
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$$\epsilon_{w} = 9\epsilon_{0}$$
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$$\epsilon_{w} = 81\epsilon_{0}$$
Explanation
The relative permittivity is defined by:
$$\epsilon_{rel} = \dfrac{\epsilon_{w}}{\epsilon_{0}}=81$$
Hence, $$\epsilon_{w}=81\times\epsilon_{0}$$
Find the charge enclosed in the cylinder.
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$$1.1\times10^{-6}$$
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$$1.1\times10^{-9}$$
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$$1.1\times10^{-11}$$
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$$1.1\times10^{-12}$$
Explanation
Since $$\overrightarrow{E}$$ and $$\overrightarrow{dS}$$ are in the same direction, so; $$\Phi \overrightarrow{E}.\overrightarrow{dS}=\dfrac{Q_{enclosed}}{\varepsilon _{0}}$$
$$\overrightarrow{E}\times $$ Cross Sectional area $$=\dfrac{Q_{enclosed}}{\varepsilon _{0}}$$
So, charge enclosed $$=50\times 25\times 10^{-4}\times 8.85\times 10^{-12}$$
$$=11062.5\times 10^{-16}\:C$$
$$=1.1\times 10^{-12}\: C$$
The electric field between the plates of two oppositely charged plane sheets of charge density $$'\sigma'$$ is :
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$$+ \dfrac {\sigma}{2\epsilon_{0}}$$
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$$- \dfrac {\sigma}{2\epsilon_{0}}$$
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$$\dfrac {\sigma}{\epsilon_{0}}$$
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Zero
Explanation
The electric field generated by charged plane sheet is uniform and not dependent on position.
It is given by: $$E=\dfrac{\sigma}{2 \epsilon_0}$$
Now, electric field between two opposite charged plane sheets of charge density $$\sigma$$ will be given by:
$$E= \dfrac{\sigma}{2 \epsilon_0} - \dfrac{(- \sigma)}{2 \epsilon_0} = \dfrac{\sigma}{\epsilon_0}$$
The unit of permittivity is :
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$${ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }$$
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$$N{ m }^{ 2 }{ C }^{ -2 }$$
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$$H{ m }^{ -1 }$$
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$$N{ C }^{ 2 }{ m }^{ -2 }$$
Explanation
$$\textbf{Step 1: Finding formula using permitivity }\epsilon_0$$
By Coulomb's law, the force between two charges $$F=\dfrac{q_1q_2}{4\pi \epsilon_0 r^2}$$
$$\therefore\ \ \epsilon_0=\dfrac{q_1q_2}{4\pi F r^2}$$
$$\textbf{Step 2: Finding Units of permitivity using above formula}$$
Substituting the SI unit of all quantities on RHS of equation $$(1)$$.
We get, the unit of
permittivity $$(\epsilon_0)=\dfrac{C.C}{N.m^2}=C^2N^{-1}m^{-2}$$
Hence option $$(A)$$ is correct
Electric charge $$q$$ , $$q$$ and $$-2q$$ are placed at the corners of an equilateral triangle $$ABC$$ of side $$L$$. The magnitude of electric dipole moment of the system is
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$$qL$$
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$$\sqrt2qL$$
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$$\sqrt3qL$$
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$$4qL$$
Explanation
$$\textbf{Step 1: Breaking the charges} $$
In the given situation, we do not se any dipole directly. But we can make dipoles by breaking the charges.
So, We Break the $$-2q$$ charge in two $$-q$$ charges, and we get
two dipoles as shown in the figure.
Dipole moment of both dipoles $$ P = q L $$
$$\textbf{Step 2: Calculation of net dipole moment} $$
Since dipole is vector quantity, so we add the two dipoles vectorially to get the resultant dipole moment.
Both dipoles are at angle $$ \theta = 60^{\circ} $$
So,
Resultant $$P_{\text{net}} = \sqrt{P^2 + P^2 + 2(P)(P) \cos 60} $$
$$= \sqrt{3P^2} $$ $$= \sqrt{3}P $$
$$P_{\text{net}} = \sqrt{3} (q \times L) $$
Hence, Option $$(C)$$ is correct.
Which of the following is not a property of electric lines of force?
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Lines of force start from positive charge and terminate at negative charge.
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Lines of force always intersect.
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The tangent to a line of force at any point gives the direction of the electric field(E) at that point.
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The number of lines per unit area, through a plane at right angles to the lines, is proportional to the magnitude of elctric field(E).
Explanation
Properties:
i) Lines of force start from positive charge and terminate at negative charge.
ii) Lines of force never intersect.
iii) The tangent to a line of force at any point gives the direction of the electric field(E) at that point.
iv) The number of lines per unit area, through a plane at right angles to the lines, is proportional to the magnitude of E. This means that, where the lines of force are close together, E is large and where they are far apart, E is small.
v) Each unit positive charge gives rise to $$\displaystyle\frac{1}{\varepsilon_o}$$ lines of force in free space. Hence number of lines of force originating from a point charge $$q$$ is N$$=\displaystyle\frac{q}{\varepsilon_o}$$ in free space.
The electric field outside the plates of two oppositely charged plane sheets of charge density $$\sigma $$ is :
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$$+\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } $$
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$$-\dfrac { \sigma }{ 2{ \varepsilon }_{ 0 } } $$
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Zero
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$$\dfrac { \sigma }{ { \varepsilon }_{ 0 } } $$
Explanation
The electric field due to charge plate is given by $$E=\dfrac{\sigma}{2\epsilon_0}$$
The field lines outside the plates are equal and opposite so the field outside plates will be zero. i.e $$E_{out}=\dfrac{\sigma}{2\epsilon_0}-\dfrac{\sigma}{2\epsilon_0}=0$$
The field in between the plates will be
$$E_{in}=\dfrac{\sigma}{2\epsilon_0}+\dfrac{\sigma}{2\epsilon_0}=\dfrac{\sigma}{\epsilon_0}$$
Above an infinitely large plane carrying charge density $$\sigma$$, the electric field points up and is equal to $$\dfrac {\sigma}{2\epsilon_{0}}$$. What is the magnitude and direction of the electric field below the plane?
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$$\sigma / 2\epsilon_{0}$$, down
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$$\sigma / 2\epsilon_{0}$$, up
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$$\sigma / \epsilon_{0}$$, down
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$$\sigma / \epsilon_{0}$$, up
Explanation
Electric field due to an infinite plane sheet of charge can be found using the gauss law. Consider a cylindrical gaussian surface as shown in the figure.
Using Gauss Law,
$$2E \Delta A = \cfrac{\sigma \Delta A}{\varepsilon_o}$$
$$E = \cfrac{\sigma}{2 \varepsilon_o}$$
Hence, electric field is $$\cfrac{\sigma}{2 \varepsilon_o}$$ downwards.
A girl brings a positively charged rod near a thin neutral stream of water flowing from a tap. She observes, that the water stream, bends towards her. Instead, if she is to bring a negativity charged rod near to the stream, it will :
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bend in the same direction
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bend in the opposite direction
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not bend at all
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bend in the opposite direction above and below the rod
Explanation
Water is dipolar in nature. Hence when water enters an electric field, the molecules rearrange themselves.
The case 1 in the figure, where a positively charged rod is brought near the water,
the negative charge moves towards the rod and positive charge moves away from the rod.
Hence, the attractive force is greater than the repulsive force and as a result, the water stream bends towards the rod.
To the contrast, in case 2, when a negatively charged rod is brought near the water, the molecules rearrange such that positive charge moves towards the rod and the negative charge moves away from it as shown in the figure.
In this case also, the attractive forces dominate the repulsive forces, and hence, the stream still bends towards the rod.
The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $$150\ N/C$$, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be (approximately):
$$\left[Given\ { \varepsilon }_{ 0 }=8.85\times { 10 }^{ -12 }{ C }^{ 2 }/{ N-m }^{ 2 },{ R }_{ E }=6.37\times { 1 }0^{ 6 }m \right]$$
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$$+\ 670\ kC$$
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$$-\ 670\ kC$$
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$$-\ 680\ kC$$
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$$+\ 680\ kC$$
Three point charges of $$+2q$$, $$+2q$$ and $$-4q$$ are placed at the corners $$A$$, $$B$$ and $$C$$ of an equilateral triangle $$ABC$$ of side '$$x$$'. The magnitude of the electric dipole moment of this system is :
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$$2\sqrt { 3 } qx$$
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$$2 qx$$
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$$3 qx$$
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$$3\sqrt { 2 } qx$$
Explanation
The charge arrangement can be simplified as two dipoles, one on $$CA$$ and the other on $$CB$$.
Dipole moment on $$CA = 2qx$$ in the direction of CA
Dipole moment on $$CB = 2qx$$ in the direction of CB
Angle between them = $$60^o$$
Net dipole moment = Vector sum of both the dipole moments
Magnitude = $$\sqrt{m_1^2+m_2^2+2m_1m_2\textrm{cos}(60^o)} = \sqrt{3\times(2qx)^2} = 2\sqrt{3}qx$$
The relation between the intensity of the electric field of an electric dipole at a distance $$r$$ from its centre on its axis and the distance $$r$$ is
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$$E\propto \dfrac {1}{r}$$
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$$E\propto \dfrac {1}{r^{2}}$$
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$$E\propto \dfrac {1}{r^{4}}$$
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$$E\propto \dfrac {1}{r^{3}}$$
Explanation
Electric field due to an electric dipole at axial point
$$E_{axial} = \dfrac {2kpr}{(r^{2} - l^{2})^{2}}$$
If $$r > > 1$$, then $$E_{a} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {2p}{r^{3}}\Rightarrow E\propto \dfrac {1}{r^{3}}$$
(directed from $$-q$$ to $$+q$$).
There exists an electric field of $$1$$N$$/$$C along Y direction. The flux passing through the square of $$1$$m placed in XY plane inside the electric field is?
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$$1.0Nm^2/C$$
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$$10.0Nm^2/C$$
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$$2.0Nm^2/C$$
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Zero
Explanation
The flux passing through the square of $$1$$m placed in XY plane inside the electric field is zero because by Gauss theorem we can say here closed circuit not formed.
The electric field intensity at a point near and outside the surface of a charged conductor of any shape is '$${ E }_{ 1 }$$'. The electric field intensity due to uniformly charged infinite thin plane sheet is '$${ E }_{ 2 }$$'. The relation between '$${ E }_{ 1 }$$' and '$${ E }_{ 2 }$$' is :
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$$2{ E }_{ 1 }={ E }_{ 2 }$$
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$${ E }_{ 1 }={ E }_{ 2 }$$
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$${ E }_{ 1 }=2{ E }_{ 2 }$$
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$${ E }_{ 1 }=4{ E }_{ 2 }$$
Explanation
Electric field outside the conductor $$E_1 = \dfrac{\sigma}{\epsilon_o}$$
Electric field due to uniformly charged infinitely thin plate $$E_2 = \dfrac{\sigma}{2\epsilon_o}$$
$$\implies$$ $$E_1 = 2E_2$$
Two infinite sheet carry equal and opposite uniform charge of densities $$\pm \sigma$$. The electric field in the free space between the two sheets will be :
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$$\dfrac {\sigma}{\epsilon_{0}}$$
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$$\dfrac {\sigma}{2\epsilon_{0}}$$
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$$\dfrac {2\sigma}{\epsilon_{0}}$$
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Zero
Explanation
Electric field at any point near single infinite charge carrying sheet is $$\sigma/2\epsilon_0$$. Direction is away from the sheet, if $$\sigma$$ is positive and towards the sheet, if it is negative.
So due to two infinite sheets of charge, total electric field at any point between the sheets is,
$$\dfrac{\sigma}{2\epsilon_0} - \dfrac{-\sigma}{2\epsilon_0} = \dfrac{\sigma}{\epsilon_0}$$
A cylinder of radius r and length l is placed in a uniform electric field of intensity E acting parallel to the axis of the cylinder. The total flux over curved surface area is:
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$$2\pi rE$$
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$$\left(\displaystyle\frac{2\pi}{l}\right)E$$
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$$2\pi rlE$$
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$$\displaystyle\frac{E}{2\pi rl}$$
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zero
Explanation
Total flux emerging from the curved surface of the cylinder,
$$\phi =E\cdot ds=Eds\cos\theta$$
Here, $$\theta =90^o$$
$$\phi =Eds\cos 90^o$$
$$\phi =0$$.
Two identical metal spheres charged with $$+12\mu C$$ and $$-8\mu C$$ are kept at certain distance in air. They are brought into contact and then kept at the same distance. The ratio of the magnitudes of electrostatic forces between them before and after contact is :
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$$12 : 1$$
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$$8 : 1$$
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$$24 : 1$$
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$$4 : 1$$
Explanation
Before contact
Initial force $$F_i=\dfrac{K{12}{(-8)}}{l^2} = \dfrac{96K }{l^2}$$
When brought together , the charg
e on each sphere is $$\dfrac{12-8}{2}=2$$
$$F_f=\dfrac{K4}{l^2}$$
$$\dfrac{F_i}{F_0}=24:1$$
Electric field produced due to an infinitely long straight uniformly charged wire at perpendicular distance of $$2 cm$$ is $$3\times { 10 }^{ 8 }N{ C }^{ -1 }$$. Then, linear charge density on the wire is _____________.
$$\left( K=9\times { 10 }^{ 9 }SI\quad unit \right) $$
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$$3.33\dfrac { \mu C }{ m } $$
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$$333\dfrac { \mu C }{ m } $$
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$$666\dfrac { \mu C }{ m } $$
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$$6.66\dfrac { \mu C }{ m } $$
Explanation
Given, $$r=2\times { 10 }^{ -2 }m$$,
$$E=3\times { 10 }^{ 8 }{ N }/{ C }$$
$$K=9\times { 10 }^{ 9 }{ N{ m }^{ 2 } }/{ { C }^{ 2 } }$$
The linear charge density $$I=\dfrac { E\cdot r }{ 2K } $$
$$\Rightarrow I=\dfrac { 3\times { 10 }^{ 8 }\times 2\times { 10 }^{ -2 } }{ 2\times 9\times { 10 }^{ 9 } } $$
$$\Rightarrow I=\dfrac { 1 }{ 3 } \times { 10 }^{ -3 }{ C }/{ m }$$
$$\Rightarrow =0.333\times { 10 }^{ -3 }{ C }/{ m }$$
$$=333\times { 10 }^{ -6 }{ C }/{ m }$$
$$=333{ \mu C }/{ m }$$
The electrostatic force between two point charges is directly proportional to the
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Sum of the charges
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Distance between the charges
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Permittivity of the medium
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Square of the distance between the charges
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Product of the charges
Explanation
By Coulomb's law
$$F=\frac {1}{4\pi\varepsilon _0}\cdot \frac {q_1q_2}{r^2}$$
So, the electrostatic force between two point charges is directly proportional to the product of the charge.
The black shapes in the figure are closed surfaces. The electric field lines are in red. For which case the net flux through the surfaces is non-zero?
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In all cases net flux is non-zero
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Only (c) and (d)
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Only (a) and (b)
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Only (b), (c) and (d)
Explanation
As charge is enclosed in a, b, so by Gauss's Law there will be a flux enclosed by the given surface.
Eight point charges (can be assumed as small spheres uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube .The electric flux through square surface ABCD of the cube is
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$$\dfrac{q}{24 \in_0}$$
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$$\dfrac{q}{12 \in_0}$$
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$$\dfrac{q}{6 \in_0}$$
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$$\dfrac{q}{8 \in_0}$$
Explanation
At each vertex, we have effective charge of $$q_v=q/8$$
So total charge enclosed by the cube is, $$8q_v=q$$
By Gauss law, total electric flux through the cube is, $$\phi=\dfrac{q}{\epsilon_o}$$
By symmetry, flux through each face of the cube would be equal.
So, flux trough ABCD is, $$\dfrac{\phi}{6}=\dfrac{q}{6\epsilon_o}$$
Option C is correct.
The surface charge density on a copper sphere is $$\sigma$$. So the intensity of electric field on its surface will be:
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$$\sigma$$
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$$\dfrac{\sigma}{2}$$
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$$\dfrac{\sigma}{2 \varepsilon o}$$
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$$\dfrac{\sigma}{\varepsilon o}$$
Explanation
The surface charge density on a copper sphere is $$\sigma $$
therefore$$,$$ the intensity of electric field on its surface will be$$-$$ $$\frac{\sigma }{{{ \in _0}}}$$
Hence,
option $$(D)$$ is correct answer.
The electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius $$10cm$$ surrounding the total charge is $$20Vm$$. The flux over a concentric sphere of radius $$20cm$$ will be
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$$20V.m$$
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$$25V.m$$
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$$40V.m$$
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$$200V.m$$
Explanation
According to Gauss's law total flux coming out of a closed surface enclosing charge
q
q
is given by
ϕ
=
∮
E
.
d
S
=
q
ε
0
ϕ=∮E→.dS→=qε0
From this expression, it is clear the total flux linked with a closed surface only depends on the enclosed charge and independent of the shape and size of the surface.
ϕ
=
∮
E
.
d
S
=
q
ε
0
=
20
V
m
ϕ=∮E→.dS→=qε0=20Vm
this
q
ε
0
qε0
is constant as long as the enclosed charge is constant
The flux over a concentric sphere of radius
20
c
m
=
20
V
m
Let the electrostatic field E at distance r from a point charge q not be an inverse square but, instead an inverse cubic, e.g., $$\vec{E}=k\frac{q}{r^3}\hat{r}$$. Here k is a constant. Consider the following two statements.
(i) Flux through a spherical surface enclosing the charge is $$\Phi =q_{enclosed}/\epsilon_0$$
(ii) A charge placed inside uniformly charged shell will experience a force.
Choose the correct option.
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Only (i) is valid
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Only (ii) is valid
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Both (i) and (ii) are invalid
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Both (i) and (ii) are valid
Explanation
Flux enclosed is not constant since $$\int E.ds$$ over a surface of radius r is not constant and is inversely proportional to r. hence (i) is not valid.
Also $$F = Eq$$, initially distance $$r$$ is inverse square in electric field but now distance r is changed to inverse cube in electric field. hence electric field is changed from initial condition, hence a charge will experience a force. hence (ii) is valid.
If a mass of $$20\ g$$ having charge $$3.0\ mC$$ moving with velocity $$20\ ms^{-1}$$ enters a region of electric field of $$80\ NC^{-1}$$ in the same direction as the velocity of mass, then the velocity of mass after $$3s$$ in the region will be
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$$40\ ms^{-1}$$
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$$44\ ms^{-1}$$
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$$56\ ms^{-1}$$
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$$80\ ms^{-1}$$
Explanation
As the mass is moving in the electric field, then,
$$ma = qE$$
$$a = \dfrac {qE}{m} = \dfrac {3\times 10^{-3} \times 80}{20\times 10^{-3}} = 12ms^{-2}$$
By using, $$v = u + at$$
$$v = 20 + 12\times 3 = 56\ ms^{-1}$$.
A charge $$Q$$ is situated at the centre of a cube. The electric flux through one of the faces of the cube is
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$$Q/ \epsilon_{0}$$
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$$Q/ 2\epsilon_{0}$$
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$$Q/ 4\epsilon_{0}$$
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$$Q/ 6\epsilon_{0}$$
Explanation
The electric flux through the whole cube:
$$ \phi = \cfrac{Q}{\varepsilon_0}$$
Flux through each face will be one sixth of the total:
$$ \phi = \cfrac{Q}{6\varepsilon_0}$$
The field at a distance $$r$$ from a long straight wire of charge per unit length $$\lambda$$ is:
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$$k \dfrac {\lambda}{r^{2}}$$
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$$k \dfrac {\lambda}{r}$$
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$$k \dfrac {\lambda}{2r}$$
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$$k \dfrac {2\lambda}{r}$$
Explanation
The field at a distance $$=r$$ a long straight wire of charge per unit length $$\lambda $$.
then the field $$V=k\dfrac { 2\lambda }{ r } $$
Two point charges $$-q$$ and $$+q/2$$ are situated at the origin and the point $$(a,0,0)$$ respectively. The point along the x-axis, where the electric field vanishes is
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$$x=\sqrt { 2 } a$$
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$$x=\cfrac { a }{ \sqrt { 2 } } $$
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$$x=\cfrac { \sqrt { 2 } a }{ \sqrt { 2 } -1 } $$
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$$x=\cfrac { \sqrt { 2a } }{ \sqrt { 2 } +1 } $$
Explanation
The given situation can be shown as given
Suppose the field vanished at a distance $$x$$, we have
$$\cfrac { kq }{ { x }^{ 2 } } =\cfrac { kq/2 }{ { (x-a) }^{ 2 } } $$
$$\Rightarrow 2{ (x-a) }^{ 2 }={ x }^{ 2 }$$
$$\sqrt { 2 } (x-a)=x\quad $$
$$\Rightarrow \sqrt { 2 } x-x=\sqrt { 2 } a$$
$$\Rightarrow (\sqrt { 2 } -1)x=\sqrt { 2 } a$$
$$\quad \Rightarrow x=\cfrac { \sqrt { 2 } a }{ \sqrt { 2 } -1 } $$
A flat, square surface with sides of length L is described by the equations. $$x = L, 0 \leq y \leq L, 0 \leq z \leq L$$
The electric flux through the square due to a positive point charge q located at the origin($$x = 0, y = 0, z=0 $$) is
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$$\dfrac{q}{4\varepsilon_o}$$
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$$\dfrac{q}{6\varepsilon_o}$$
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$$\dfrac{q}{24\varepsilon_o}$$
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$$\dfrac{q}{48\varepsilon_o}$$
Explanation
This can be done by assuming the cube having x,y,z direction.
$$\phi$$(Flux through symmetric cube)$$=\dfrac{q}{\varepsilon_0}$$
Cube has six faces,So flux through one face $$=\dfrac{q}{6\varepsilon_0}$$
Three charges are arranged at the vertices of an equilateral triangle of side $$l$$. The dipole moment of the combination is
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Zero
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$$\sqrt { 3 } ql$$
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$$ql/\sqrt { 3 } $$
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$$\sqrt { 2 } ql$$
Explanation
The components of $${ \vec { P } }_{ 1 }$$ & $${ \vec { P } }_{ 2 }$$ along the $$x$$ -axis will cancel out each other while the component along $$y$$ -axis will add up
$$\left| { \vec { P } }_{ net } \right| =\left[ \left( ql \right) \cos { { 30 }^{ ° } } \right] \times 2$$
$$=\left( ql \right) \cfrac { \sqrt { 3 } }{ 2 } \times 2$$
$$\left| { \vec { P } }_{ net } \right| =\sqrt { 3 } ql$$
Hence option (B) is correct
In a Neon discharge tube $$2.9\ \times \ { 10 }^{ 18 } \ ({ Ne }^{ + })$$ ions move to the right each second, while $$1.2\ \times \ { 10 }^{ 18 }$$ electrons move to the left per second; electron charge is $$1.6\ \times \ { 10 }^{ -19 } C$$. The current in the discharge tube is:
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$$1 A$$ toward right
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$$0.66 A$$ toward right
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$$0.66 A$$ toward left
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Zero
Explanation
Given, rate of flow of change, of
$$Ne^+$$ ions, $$I_1=2.9\times 10^{18}\times 1.6\times 10^{-19}$$ [No. of ions $$\times $$ change of each ions]
$$=0.464A$$
of $$e^-$$, $$I_2=1.2\times 10^{18}\times 1.6\times 10^{-19}$$ [conventional current is taken opposite to flow of $$e^-$$]
$$=0.192A$$
Thus, net current
$$I=I_1+I_2$$
$$=0.464+0.192$$
$$\Rightarrow I=0.656A$$ towards right.
An electric dipole has a pair of equal and opposite point charges + Q and Q separated by a distance 2x. The axis of the dipole is defined as
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the line joining positive charge to -ve charge
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the line making angle of $$45^{o}$$ with line joining two charges
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perpendicular to the line joining the two charges drawn at the centre and pointing upward direction.
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perpendicular to the line joining the two charges drawn at the centre and pointing downward direction.
Explanation
An electric dipole has a pair of equal and opposite $$+Q$$ and $$-Q$$ distance $$=2x$$. In this context, axis of an electric dipole is always directed from $$(-Ve)$$ charge.
A hemisphere shell is uniformly charged positively. The electric field at a point on a diameter away from the centre (inside the boundary of hemisphere shell) is directed
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perpendicular to the diameter
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parallel to the diameter
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at an angle tilted towards the diameter
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at an angle tilted away from the diameter
Explanation
It has to be perpendicular as when you join the other half to make complete sphere, only this orientation of the electric field will give a vector sum equal to $$0$$, inside the sphere.
Two copper spheres, $$A$$ and $$B$$, are identical in all respect but A carries a charge of $$-3 \mu C$$ whereas $$B$$ Is charged to $$+1 \mu C$$. The two spheres are brought together until they touch and then separated by some distance. Which of the following statements is true concerning the electrostatic force $$F$$ between the spheres?
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$$F = 0$$ as one of the spheres is uncharged
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$$F = 0$$ as both the spheres are uncharged
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$$F$$ is attractive
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$$F$$ is repulsive.
Explanation
Charge on $$A^{+}$$ sphere $$=-3$$ $$uC$$
Charge on $$B$$ sphere $$=+1$$ $$uC$$
When they are connected the charge get redistributed, i.e.
$$\Rightarrow$$ Charge on both sphere $$=\dfrac{-3+1}{2}$$ $$uC=-1$$ $$uC$$
So charge on either sphere $$=-1$$ $$uC$$
Therefore, After connecting charge on sphere $$A=-1$$ $$uC$$
After connecting charge on sphere $$B=-1$$ $$uC$$
Therefore, there will be a repulsive force between the sphere because of like charges.
Electric field intensity at points in between and outside two thin separated parallel sheets of infinite dimension with like charges of same surface charge density ($$\sigma$$) are ______ and ______ respectively
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$$\sigma /{ \epsilon }_{ 0 },\sigma /{ \epsilon }_{ 0 }$$
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$$0,\sigma /{ \epsilon }_{ 0 }$$
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$$0,0$$
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$$\sigma /{ \varepsilon }_{ 0 },0$$
Explanation
The electric field inetensity due to the long sheet of charge is given by:
$${ E }_{ 1 }={ E }_{ 2 }=\cfrac { \sigma }{ 2{ \epsilon }_{ 0 } } $$
The electriic field intensity between plates
$$E_{in}={ E }_{ 1 }-{ E }_{ 2 }$$
$$\Rightarrow E=0$$
The electric field intensity outside plates
$$E_{out}={ E }_{ 1 }+{ E }_{ 2 }$$
$$\Rightarrow E=\cfrac { \sigma }{ { \epsilon }_{ 0 } } $$
In a certain region of space,electric field is along the z-direction throughout. The magnitude of electric field is however not constant, but increases uniformly along the positive z-direction at the rate of $${ 10 }^{ 5 }N\quad { C }^{ -1 }{ m }^{ -1 }$$. Torque experienced by the system is:
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$${ 10 }^{ 2 }N$$
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$${ 10 }^{ -2 }N$$
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zero
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$${ 10 }^{ 3 }N$$
Explanation
The force on the dipole is along negative direction of z-axis, so $$\theta ={ 180 }^{ o }$$
$$\therefore$$ Torque on dipole, $$\tau =PE\sin { { 180 }^{ o } } =0$$
The force between two charges in different media are different because.
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Different media have different permittivity's
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Different media have different viscosities
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Different media have different densities
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Different media have different permeabilities
Explanation
Force between two charges depends upon the permittivity of the media and the different media has different
permittivity. So,
The force between two charges in different media are different.
The nucleus of helium atom contains two protons that are separated by distance $$3.0\times {10}^{-15}m$$. The magnitude of the electrostatic force that each proton exerts on the other is:
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$$20.6N$$
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$$25.6N$$
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$$15.6N$$
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$$12.6N$$
Explanation
Charge of proton is $${ q }_{ p }=1.6\times { 10 }^{ -19 }C$$
Distance between the ptoton is $${ r }=3\times { 10 }^{ -15 }m$$
The magnitude of electrostatic force between protons is
$${ F }_{ e }=\cfrac { { q }_{ p }{ q }_{ p } }{ 4\pi { \varepsilon }_{ 0 }{ r }^{ 2 } } =\cfrac { 9\times { 10 }^{ 9 }\times 1.6\times { 10 }^{ -19 }\times 1.6\times { 10 }^{ -19 } }{ { \left( 3\times { 10 }^{ -15 } \right) }^{ 2 } } =25.6N$$
The solid angle subtended by the periphery of an area $$1 cm^2$$ at a point situated symmetrically at a distance of 5 cm from the area is
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$$2 \times 10^{-2}$$ steradian
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$$4 \times 10^{-2}$$ steradian
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$$6 \times 10^{-2}$$ steradian
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$$8 \times 10^{-2}$$ steradian
Explanation
Given data,
$$Area=1cm^{2}$$
$$Distance=5cm$$
$$Solid \ angle \ \Omega=?$$
We know the formula
Solid angle $$\Omega=\dfrac{Area}{(Radial \ distance)^{2}}$$
$$=\dfrac{1cm^{2}}{(5cm^{2})}=\dfrac{1}{25}$$
$$=4\times 10^{-2}steradian$$
The solid angle is for $$3-D$$ figure like spere, cone, etc and plane angle is for plane objects or $$2-D$$ figure like circle,arc etc.
The constant $$k$$ in Coulomb's law depends upon
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nature of medium
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system of units
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intensity of charge
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both (a) and (b)
Explanation
The value of $$k=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } =8.854\times { 10 }^{ -12 }{ C }^{ 2 }{ N }^{ -1 }{ m }^{ -2 }$$
where $${ \varepsilon }_{ 0 }$$ is permittivity of free space
Which of the following figure represents the electric field lines due to the combination of one positive and one negative charge?
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0%
0%
0%
Explanation
The electric force in between the combination of one positive and one negative charge used to be attractive in nature and electric field lines between them originate from positive charge and end at the negative charge.
Hence, the correct option is $$(A)$$
The unit of polarizability of the molecule is __________.
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$$C^2m^1N^{-1}$$
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$$C^2m^{-1}N^1$$
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$$C^{-2}m^1N^{-1}$$
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$$C^2m^{-1}N^{-1}$$
Explanation
Polarizability of the molecule is defined as the electric dipole moment induced in the molecule per unit incident electric field.
Polarizability $$\alpha = \dfrac{P}{E}$$
Units $$\alpha = \dfrac{C \ m}{N/C} = C^2 m N^{-1}$$
Figure shows the electric field lines around three point charges $$A,B$$ and $$C$$. Which region or regions of the figure could the electric field be zero?
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Near A
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Near B
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Near C
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Nowhere
Explanation
From the figure, it is clear that, there cannot be a neutral point between a positive and a negative charge. Neutral point can be only there between two like charges i.e., $$A$$ and $$C$$. As magnitude of charge $$A$$ is smaller than the magnitude of charge $$C$$, therefore, neutral point would lie closer to the charge $$A$$.
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Practice Class 12 Medical Physics Quiz Questions and Answers
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