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CBSE Questions for Class 12 Medical Physics Electrostatic Potential And Capacitance Quiz 6 - MCQExams.com
CBSE
Class 12 Medical Physics
Electrostatic Potential And Capacitance
Quiz 6
A soap bubble of radius $$\sqrt 7$$cm is blown. Work done in this process is used to store energy in capacitor at potential of $$1$$V using transducer. Charge stored is ________.
$$(T=30 dyne/cm)$$
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$$5.28\mu C$$
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$$10.56\mu C$$
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$$21.12\mu C$$
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$$21.12 mC$$
Explanation
Change in energy of the soap bubble $$=\Delta A\times T$$.
Change in Area $$=2\times 4\pi { R }^{ 2 }$$, (as the soap bubble has two surfaces).
Change in Energy $$=2\times 4\pi \times 7\times 30\ ergs$$
Thus, Energy of the capacitor $$=$$
work done on the bubble.
$$\dfrac { 1 }{ 2 } QV=2\times 4\pi \times 7\times 30\times { 10 }^{ -9 } $$
$$Q=10.56 \mu C$$
Two charges of equal magnitude 'q' are placed in air at a distance '$$2a$$' apart and third charge '$$-2q$$' is placed at midpoint. The potential energy of the system is ________.
($$\epsilon_0=$$ permittivity of free space)
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$$\displaystyle -\frac{q^2}{8\pi \epsilon_0a}$$
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$$\displaystyle -\frac{3q^2}{8\pi \epsilon_0a}$$
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$$\displaystyle -\frac{5q^2}{8\pi \epsilon_0a}$$
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$$\displaystyle -\frac{7q^2}{8\pi \epsilon_0a}$$
Explanation
The system of 3 charges can be thought of as three pairs of charges and the potential energy of each pair can be found individually. The total potential energy will be the sum of these energies.
Therefore, the potential energy of the system is given by:
$$ U=\dfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } [\dfrac { { q }^{ 2 } }{ 2a } -\dfrac { { 2q }^{ 2 } }{ a } -\dfrac { { 2q }^{ 2 } }{ a } ]=\dfrac { -7{ q }^{ 2 } }{ 8\pi { \varepsilon }_{ 0 }a } $$
Figure shows three spherical and equipotential surfaces $$A$$, $$B$$ and $$C$$ around a point charge $$q$$. The potential difference $${ V }_{ A }-{ V }_{ B }={ V }_{ B }-{ V }_{ C }$$. If $${ t }_{ 1 }$$ and $${ t }_{ 2 }$$ be the distances between them, then :
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$${ t }_{ 1 }={ t }_{ 2 }$$
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$${ t }_{ 1 }>{ t }_{ 2 }$$
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$${ t }_{ 1 }<{ t }_{ 2 }$$
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$${ t }_{ 1 }\le { t }_{ 2 }$$
Explanation
The potential around a point charge is $$V(r) = \displaystyle \frac{q}{4\pi\epsilon_0r}$$
Let $$a,b,c$$ be the radii of $$A,B,C$$ respectively.
Thus, $$t_1 = b-a$$ and $$t_2 = c-b$$
$$V_A - V_B = \displaystyle \frac{q}{4\pi\epsilon_0a} - \frac{q}{4\pi\epsilon_0b} = \frac{q\ t_1}{4\pi\epsilon_0ab}$$
$$V_B-V_C = \displaystyle \frac{q}{4\pi\epsilon_0b} - \frac{q}{4\pi\epsilon_0c} = \frac{q\ t_2}{4\pi\epsilon_0bc}$$
But, $$V_A - V_B = V_B - V_C$$
$$\Rightarrow \displaystyle \frac{V_A - V_B}{V_B - V_C} = 1\Rightarrow \frac{t_1}{ab}\times \frac{bc}{t_2}=1\Rightarrow\frac{t_1}{t_2} = \frac{a}{c}$$
As $$a<c$$, we have $$t_1 < t_2$$
High frequency capacitor offers
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more resistance
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less resistance
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zero resistance
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None of these
Explanation
According to formula,
$${X}_{C}=\cfrac{1}{\omega C}=\cfrac{1}{2\pi C}$$
When $$f$$ is high, $${X}_{C}$$ is low
When a dielectric slab is introduced between the two plates of condenser then its capacity ___________.
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remains constant
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increases
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decreases
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may increase or decrease depending on the material of dielectric slab
Explanation
As the dielectric slab is introduced there is some charge distribution in the slab and because of this the electric field between the two plates is decreased, due to which the capacitor can hold more charge. Thus, the capacity to hold charge of the capacitor is increased.
A parallel plate condenser with oil (dielectric constant $$2$$) between the plates has capacitance $$C$$. If the oil is removed, the capacitance of capacitor becomes :
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$$\sqrt {2C}$$
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$$2C$$
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$$\cfrac{C}{\sqrt {2}}$$
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$$\cfrac{C}{2}$$
Explanation
The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is
$$C=\cfrac { K{ \varepsilon }_{ 0 }A }{ d } ........(i)$$
Where $${ \varepsilon }_{ 0 }=$$ electric permitivity of free space
$$K=$$dielectric constant
$$A=$$ area of each plate of capacitor
$$d=$$distance between two plates
When dielectric (oil) is removed, so capacitance
$${C}_{0}=\cfrac { { \varepsilon }_{ 0 }A }{ d } ............(ii)$$
Comparing Eqs. $$(i)$$ and $$(ii)$$ we get
$$C=K{C}_{0}$$
$$\Rightarrow$$ $${C}_{0}=\cfrac{C}{K}=\cfrac{C}{2}$$ ($$\because K=2$$)
If dielectric is inserted in charged capacitor (battery removed ), then quantity that remains constant is.
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Capacitance
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Potential
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Intensity
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Charge
Explanation
Variation of different variables $$(Q, C, V, E$$ and $$U)$$ of parallel plate capacitor when dielectric $$(K)$$ is introduced when battery is removed is
$$C'=KC$$ $$E'=E/K$$
$$Q'=Q$$ $$U'=U/K$$
$$V'=V/K$$
The capacity of parallel plate capacitor increases with the
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Increase of its distance
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Increases of its area
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Decreases of its area
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None of the above
Explanation
Hint: Use the formula of capacitance for parallel plate capacitor
Step1: Expression of capacitance
A parallel plate capacitor with a dielectric slab of dielectric constant $$3$$, filling the space between the plates, is charged to a potential $$ V$$. The battery is then disconnected and the dielectric slab is withdrawn. It is then replaced by another dielectric slab of dielectric constant $$2$$. If the energies stored in the capacitor before and after the dielectric slab is changed are $$\displaystyle { E }_{ 1 }$$ and $$\displaystyle { E }_{ 2 }$$, then $$\displaystyle { E }_{ 1 }/{ E }_{ 2 }$$ is:
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$$\displaystyle \frac { 4 }{ 9 } $$
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$$\displaystyle \frac { 2 }{ 3 } $$
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$$\displaystyle \frac { 3 }{ 2 } $$
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$$\displaystyle \frac { 9 }{ 5 } $$
Explanation
Let the charge stored by capacitor with dielectric constant $$3$$ be $$Q$$.
Thus energy stored be $$\dfrac{Q^2}{2C_1}$$
Since the charge remains the same after changing the dielectric, the new energy stored=$$\dfrac{Q^2}{2C_2}$$
$$\implies \dfrac{E_1}{E_2}=\dfrac{C_2}{C_1}=\dfrac{K_2}{K_1}=\dfrac{2}{3}$$
$$(C=\dfrac{KA\epsilon_0}{d})$$
An air filled parallel plate condenser has a capacity of $$2pF$$. The separation of the plates is doubled and the inter-space between the plates is filled with wax. If the capacity is increased to $$6$$pF, the dielectric constant of wax is.
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$$2$$
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$$3$$
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$$4$$
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$$6$$
Explanation
$$C_{old}=\cfrac{\varepsilon_oA}{d}=2pF\longrightarrow(i)$$
$$C_{new}=\cfrac{K\varepsilon_oA}{2d}=6pF\longrightarrow(ii)$$
By Equating $$(i)$$ and $$(ii)$$
$$\cfrac{K}{2}\left(2\right)pF=6pF$$
$$\Rightarrow K=6$$
Three particles, each having a charge of 10 $$\mu$$C are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is :$$\displaystyle \left(\text{Given} \frac{1}{4\pi \in_0}=9\times 10^9N-m^2C^2\right)$$
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Zero
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$$\infty$$
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27 J
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100 J
Explanation
Work done in placing first charge at one of the corners=$$0$$(no other charge present)
Work done in placing second charge at the other corner=$$\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r_{12}}$$
Work done in placing third charge=$$\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_3}{r_{13}}+\dfrac{1}{4\pi\epsilon_0}\dfrac{q_2q_3}{r_{23}}$$
Thus the total work done=
$$\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r_{12}}+\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_3}{r_{13}}+\dfrac{1}{4\pi\epsilon_0}\dfrac{q_2q_3}{r_{23}}=27J$$
Hence correct answer is option C.
In a capacitive circuit, dielectrics materials are placed between the plates of capacitors to
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0%
Speed the current flow
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Slow the current flow
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Reduce change leakage from the capacitor
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increase change leakage from the capacitor
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Increase capacitance of the capacitor
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Decrease capacitance of the capacitor
Explanation
When a dielectric material with dielectric constant K is placed in between the plates of a capacitor, the capacitance will be increased by factor K. Thus, the option E will be correct.
Two positively charged particles are placed at different positions in a cube. In which diagram is the electrical potential energy of the pair of charges the greatest?
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0%
0%
0%
0%
Explanation
The potential energy of pair of charges is $$U=\dfrac{KQQ}{r^2}$$ where r be the distance between charges.
When r is least so the potential energy will greatest. Thus for the diagram of option E , r will be least.
A student wants to increase the capacitance of a parallel-plate capacitor. How can he/she do this ?
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Using smaller plates
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Replacing the dielectric material between the plates with one that has a smaller dielectric constant
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Decreasing the voltage between the plates
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Increasing the voltage between the plates
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Moving the plates closer together
Explanation
We know that the capacitance, $$C=\dfrac{A\epsilon_0}{d}$$ where A be the area of plate and d be the separation between plates.
Also $$C=\frac{Q}{V}$$ and the capacitance is proportional to dielectric constant.
From above information we say that only option E will give correct answer.
In above shown figure, the two plates of the capacitor of charge +Q and -Q, which points lie on the same equipotential?
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1 and 2 only
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1 and 3 only
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2 and 4 only
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3 and 4 only
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1, 2, 3, and 4 all lie on the same equipotential, since the electric field is uniform
Explanation
For parallel plate capacitor, $$E=V/d$$ where d be the separation between plates. As the field E between the plates in uniform so the potential will vary linearly with distance. Since points 2 and 4 are equidistant from plates so the potential will be same at 2 and 4.
The potential difference between the plate of a capacitor is increased by $$20$$%. The energy stored on the capacitor increases by:
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$$20$$%
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$$22$$%
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$$40$$%
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$$44$$%
Explanation
Energy stored in a capacitor with potential difference $$V$$ applied across it is $$\dfrac{1}{2}CV^2=E_0$$.
When potential difference increases by 20%, $$V'=1.2V$$
$$\implies E'=\dfrac{1}{2}CV'^2=\dfrac{1}{2}C(1.2^2V^2)$$
$$\implies E'=1.2^2E_0=1.44E_0$$
Thus the energy stored increases by $$44$$%.
If the charge on a capacitor is increased by $$2$$ coulomb, the energy stored in it increases by 21%. The original charge on the capacitor is :
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$$10C$$
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$$20C$$
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$$30C$$
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$$40C$$
Explanation
According to question
Percentage increase in energy of capacitor=$$\dfrac { \dfrac { { q }_{ f }^{ 2 } }{ 2C } -\dfrac { { q }_{ i }^{ 2 } }{ 2C } }{ \dfrac { { q }_{ i }^{ 2 } }{ 2C } } \times 100=21$$ ... (i) and $${ q }_{ f }-{ q }_{ i }=2$$ ...(ii)
From (i) and (ii)
$$ 4(q_i + 1) = 0.21 (q_i)^2 $$
On solving:
Solving, we get $${ q }_{ i }=20C$$
Equipotential lines are shown, what is the approximate voltage at point P in the diagram?
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12 v
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8.0 v
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8.5 v
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6.0 v
Explanation
Voltages of all the equipotential lines are shown in the figure.
As the point P lies appoximately at the midpoint between 2nd and 3rd equipotential line, thus average potential at point P would be $$8.5$$ $$V.$$
Two parallel-plate of capacitor have charges +Q and -Q and potential difference $$\triangle V$$ due to charging, Now the capacitor is disconnected then
the potential difference and the stored electrical potential energy is:
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The potential difference decreases, and the stored electrical potential energy decreases
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The potential difference decreases, and the stored electrical potential energy increases
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The potential difference increases, and the stored electrical potential energy decreases
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The potential difference increases, and the stored electrical potential energy increases
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The potential difference decreases, and the stored electrical potential energy remains unchanged.
Explanation
When the capacitor is discharged from the battery then the charge cannot go anywhere from the capacitors and thus the energy in it remains constant.
A dielectric is inserted into a capacitor while keeping the charge constant. Identify what happens to the potential difference and the stored energy?
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The potential difference decreases and the stored energy increases
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Both the potential difference and the stored energy increases
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The potential difference increases and the stored energy decreases
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Both the potential difference and the stored energy decrease
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Both the potential difference and the stored energy remain the same
Explanation
The capacitance of the capacitor increases due to the insertion of the dielectric, $$C' = kC$$ ; where, $$k>1$$
Initial potential difference on the capacitor $$ V = \dfrac{Q}{C}$$
New potential difference $$V' = \dfrac{Q}{C'} = \dfrac{Q}{kC} =\dfrac{V}{k}$$
Thus potential difference decreases.
Also initial energy stored in the capacitor $$E = \dfrac{Q^2}{2C}$$
New energy stored $$E' = \dfrac{Q^2}{2 C'} = \dfrac{Q^2}{2 (kC)} = \dfrac{E}{k}$$
Thus energy stored in the capacitor also decreases.
How much power is used in moving a $$0.03C$$ charge from point A at a potential of $$40V$$ to point B at $$60V$$, if this takes 2 seconds?
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$$0.30W$$
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$$0.15W$$
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$$0.6W$$
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$$0.012W$$
Explanation
Given : $$q = 0.03$$ C $$V_A = 40$$ V $$V_B = 60$$ V $$t = 2$$ s
Work done in moving a charge from A to B $$W = q(V_B - V_A) = 0.03 \times (60-40) = 0.6$$ J
$$\therefore$$ Power used $$P = \dfrac{W}{t} =\dfrac{0.6}{2} = 0.3$$ $$W$$
The diagram shows equipotential lines from an unknown charge configuration. Determine the direction of the field at A :
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Up
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Down
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Left
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Right
Explanation
The electric field is given by $$E=-\dfrac{dV}{dx}$$
Hence it is in the direction of decreasing potential.
The potential at A is -5V and that to the left of it is -10V. Hence the electric field is towards left.
The diagram shows equipotential lines from an unknown charge configuration. Determine where in the diagram the field is closest to uniform :
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A
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B
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C
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D
Explanation
The electric field is given by $$E=-\dfrac{dV}{dx}$$
Hence electric field is uniform where the density of the electric field spread is uniform, and hence at point B.
What is the change in potential energy of a particle of charge +q that is brought from a distance of 3r to a distance of 2r by a particle of charge q?
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$$kq^2/r$$
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$$-kq^2/6r$$
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$$kq^2/r^2$$
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$$-kq^2//4r^2$$
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$$8kq^2/r^2$$
Explanation
As the $$+q$$ is brought near by charge $$q$$ so there is an attraction force it means q is a negative charge, equal to $$-q$$.
The electric potential energy of a system of two point charges ( $$+q$$ , $$-q$$) sepatrated by a distance $$3r$$ is given by ,
$$U=k\dfrac{q\times -q}{3r}$$ ,
when the separation is reduced to $$2r$$ then potential energy will become ,
$$U'=k\dfrac{q\times- q}{2r}$$
therefore change in potential energy is
$$\Delta U=U'-U$$
or $$\Delta U=-k\dfrac{q\times q}{2r}+k\dfrac{q\times q}{3r}$$
or $$\Delta U=kq^{2}\left(1/3r-1/2r\right)=-kq^{2}/6r$$
A dielectric is inserted into a capacitor while the charge on it is kept constant. What happens to the potential difference and the stored energy?
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The potential difference decreases and the stored energy increases
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Both the potential difference and the stored energy increase
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The potential difference increases and the stored energy decreases
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Both the potential difference and the stored energy decrease
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Both the potential difference and the stored energy remain the same
Explanation
When a dielectric is placed in an electric field then an internal electric field is produced in it due to polarisation of dielectric . This internal electric field inside the dielectric is in opposite direction of the field between plates of capacitor , as a result of this effective electric field between plates decreases , hence the potential difference between plates because ,
$$E=\Delta V/\Delta r$$ ,
where $$\Delta r $$ is the distance between plates , which is constant .
Now potential energy of a capacitor is given by ,
$$U=(1/2)q\Delta V$$ ,
charge is given constant , therefore due to decrease in potential difference $$\Delta V$$ , potential energy $$U$$ also decreases .
Calculate the change in potential energy of a particle of charge $$+q$$
that is brought from a distance of $$3r$$
to a distance of $$2r$$ in the electric field of
charge $$-q$$?
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$${ k{ q }^{ 2 } }/{ r }$$
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$${ -k{ q }^{ 2 } }/{ 6r }$$
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$${ k{ q }^{ 2 } }/{ 4{ r }^{ 2 } }$$
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$${ -k{ q }^{ 2 } }/{ 4{ r }^{ 2 } }$$
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$${ k{ q }^{ 2 } }/{ { r }^{ 2 } }$$
Explanation
Given : $$PB = 2r$$ $$PA = 3r$$
Potential at point A $$V_A = \dfrac{k q}{PA} = \dfrac{kq}{3r}$$
Potential at point B $$V_B = \dfrac{k q}{PB} = \dfrac{kq}{2r}$$
Change in potential $$\Delta V = V_B - V_A = \dfrac{kq}{2r} - \dfrac{kq}{3r} = \dfrac{kq}{6r}$$
Thus change in potential energy by moving a charge $$-q$$ from A to B $$W = -q (\Delta V) = \dfrac{-kq^2}{6r}$$
A negative charge is located in the uniform electric field between two plates (not shown).
Where in the diagram will the electron have the greatest change in potential energy before hitting the one of the plates?
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A
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B
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C
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D
Explanation
A charged species tends to move from place of higher potential energy to that of lower potential energy.
The direction of electric field is from left to right. This means that there is a positively charged plate on the left side, closest to A and farthest from B.
An electron being negatively charged plate always moves towards left here. Hence the maximum potential energy resides at B.
The diagram shows three identical metal plates A, B, and C that are set parallel to each other. Plate A is farther from B than C is from B. A source of constant EMF is connected to the system such that B is connected to the positive terminal and A and C are connected to the negative terminal.
Which of the following diagrams correctly depicts the isolines of electric potential (shown as dotted lines) in the region between the plates?
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0%
0%
0%
0%
Explanation
Since same battery has been joined between the two capacitors, the potential difference between both set of plates is same.
However the distance between A and B is more, hence the electric field between A and B is less. So the gradient between A and B must be less. Hence equipotential surfaces should be denser between B and C.
Correct answer is option B.
A capacitor loses half of its voltage every 1 seconds.
If the initial Energy it holds is $$E$$, how much energy will it have after 3 seconds?
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$$E/3$$
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$$E/8$$
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$$E/9$$
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$$E/64$$
Explanation
Energy stored in a capacitor is $$U=E=\frac{1}{2}CV^2$$
Here after 1 second, the voltage will be $$V/2$$
Thus, after 3 seconds, the voltage will be $$V/8$$
So, after 3 seconds , the energy will be $$E'=\frac{1}{2}C(V/8)^2=E/64$$
An air-filled capacitor has a capacitance of $$C^o$$. Then a dielectric with a dielectric constant of 3.0 is inserted into it and the plates are pushed toward each other so that the distance between the plates is reduced to $$\frac{1}{2}$$ the original value.
What is the new capacitance?
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$$\frac{1}{6}C^o$$
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$$\frac{2}{3}C^o$$
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$$\frac{3}{2}C^o$$
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$$6C^o$$
Explanation
Capacitance of air-filled parallel plate capacitor $$C^o =\dfrac{A\epsilon_o}{d}$$
Dielectric constant of the inserted dielectric $$K = 3.0$$
New distance between the plates $$d' =d/2$$
$$\therefore$$ New capacitance $$C' = \dfrac{KA\epsilon_o}{d'} = \dfrac{(3.0)A\epsilon_o }{d/2} = 6C^o$$
Three capacitors, $$3\mu F, 6\mu F$$ and $$6\mu F$$ are connected in series to a source of 120V. The potential difference, in volts,
across the $$3\mu F$$ capacitor will be
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0%
24
0%
30
0%
40
0%
60
Explanation
The equivalent capacitance of the two $$6\mu F$$ and $$6\mu F$$ capacitors in series is $$3\mu F$$.
Hence the potential across the two capacitors, original $$3\mu F$$ capacitor and the equivalent $$3\mu F$$ capacitor is divided equally.
Hence voltage across each of the capacitors is half of the external applied voltage, $$60V.$$
In context of a capacitor , which of the following statement is correct?
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Electric field may be uniform
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There must be some equipotential surface somewhere inside
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Electric potential must be max at surface of one of the plate in case of parallel plate capacitor
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None of the above
Explanation
Electric field at the surface of a capacitor is Zero.
$$\therefore E=0$$
Also, $$E=\cfrac { \delta V }{ \delta x } $$
$$\therefore \cfrac { \delta V }{ \delta x } =0$$
$$\therefore V$$ is maximum at the surface.
An electric dipole is kept in the origin with charges along the x axis, now choose the correct option,
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equipotential surface is on $$xy$$ plane
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equipotential surface is on $$xz$$ plane
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equipotential surface is on $$yz$$ plane
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none of the above
Explanation
Electric potential of every point lying on the $$\mathbf{YZ}$$ plane is zero as every point is at equal distance from negative and positive charge of dipole. So, the plane will be equipotential.
Therefore, (C) is correct option.
A capacitor contains two square plates with side lengths $$5.0$$ cm. The plates are separated by $$2.0$$ mm. Dry air fills the space between the plates. Dry air has a dielectric constant of $$1.00$$ and experiences dielectric breakdown when the electric field exceeds $$3.010^4 V/cm$$.
What is the magnitude of charge that can be stored on each plate before the capacitor exceeds its breakdown limit and sends a spark between the plates?
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$$6.6\times 10^{-8}C$$
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$$6.6\times 10^{-5}C$$
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$$3.3\times 10^{-7}C$$
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$$3.3\times 10^{-8}C$$
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$$8.1\times 10^{-2}C$$
Explanation
We have $$C=Q/V$$
or $$Q=CV$$ ,
for parallel plate capacitor , $$C=\varepsilon_{0} KA/d$$ ,
where $$A=$$ area of a plate ,
$$d=$$ separation between plates ,
and $$E=V/d$$ or $$V=Ed$$
therefore $$Q=\varepsilon_{0}KA/d\times Ed$$ ,
or $$Q=\varepsilon_{0}KA\times E$$
now charge on a plate $$Q$$ will be maximum when electric field $$E$$ is maximum ,
maximum value of $$E=3.0\times 10^{4} V/cm=3.0\times10^{6}V/m$$ ,
area of a plate $$A=5.0^{2}=25cm^{2}=25\times10^{-4}m^{2}$$ ,
therefore maximum charge ,
$$Q=8.854\times10^{-12}\times1\times25\times10^{-4}\times3.0\times10^{6}$$
or $$Q=6.6\times10^{-8}C$$
Electric potential is the force experienced by a unit positive charge placed at a point
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True
0%
False
Explanation
We know $$F=qE$$
where $$\overrightarrow {F}$$ is the force experienced by a charge q placed in the electric field of strength $$\overrightarrow{E}$$.
The two capacitors $$2\mu F$$ and $$6\mu F$$ are put in series, the effective capacity of the system is $$\mu F$$ is:
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$$8\mu F$$
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$$2\mu F$$
0%
$$3/2\mu F$$
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$$2/3\mu F$$
Explanation
When connected in series
$$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$$
$$\dfrac{1}{C}=\dfrac{1}{2}+\dfrac{1}{6}$$
$$C=\dfrac{3}{2}$$
When two capacitors of capacities of $$3\mu F$$ and $$6\mu F$$ are connected in series and connected to $$120\ V$$, the potential difference across $$3\mu F$$ is:
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$$40\ V$$
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$$60\ V$$
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$$80\ V$$
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$$180\ V$$
Explanation
Equivalent capacitance is C
$$\dfrac{1}{C}=\dfrac{1}{3}+\dfrac{1}{6}$$, So $$C=2\mu f$$
Now $$Q=VC=120\times 2=240\mu F$$
Now potential across $$3\mu f$$ is $$V=\dfrac{Q}{3}=240/3=80V$$
The equivalent capacitance of capacitors $$6\mu F$$ and $$3\mu F$$ connected in series is ______.
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$$3\mu f$$
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$$2\mu f$$
0%
$$4\mu f$$
0%
$$6\mu f$$
Explanation
We know t
he equivalent capacitance of capacitors
connected in series can be found by using
$$\dfrac{1}{C_{eq}}$$
$$=\dfrac{1}{C_{1}}$$
$$+\dfrac{1}{C_{2}}$$
$$+\dfrac{1}{C_{3}}+...$$
$$\dfrac{1}{C_{eq}}$$
$$=\dfrac{1}{6}$$
$$+\dfrac{1}{3}$$
$$\Rightarrow C_{eq} = \dfrac{3\times 6}{3+6} = 2\mu F $$
Therefore, B is correct option.
The amount of work done in moving a unit positive charge from infinity to a given point is known as:
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0%
Nuclear potential
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Potential energy
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Electric potential
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Gravitational potential
Explanation
Electric potential may be defined as the amount of work done in moving a unit positive charge from infinity to a given point.
$$V = \cfrac{W}{q}$$
An electric dipole is placed at the centre of a sphere, choose the correct options :
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electric field is zero at every point on the surface
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flux is zero across the surface
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no circle is present in the sphere which is equipotential
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none of the above
Explanation
A dipole contains two equal and opposite charge. So total charge inside the sphere will be zero.
By Gauss's law, the flux across a surface is depends on the charge inside the surface. As total charge is zero inside the sphere so the flux through the sphere will be zero.
As the electric field is resultant effect due to all charges so there will be field exists on the surface.
As the sphere contains two equal and opposite charges so there may be exists equipotential surface in the sphere.
Which of the following pairs are analogous?
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Energy and power
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Gravitational potential energy and electric potential energy
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Electric field and gravitational field
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None of the above
Explanation
We have to find which pairs are analogous out of four options.
Since gravitational force is analogous to electrostatic force. Similarly gravitational potential energy $$=\dfrac{Gm_{1}m_{2}}{r}$$ is analogous to electric potential energy $$=\dfrac{kq_{1}q_{2}}{r}$$
Similarly, electric field $$=\dfrac{kq_{1}q_{2}}{r^{2}}$$ is analogous to gravitational field $$=\dfrac{Gm_{1}m_{2}}{r^{2}}$$
$$K=\dfrac{1}{4\pi \varepsilon _{0}}$$, constant in coulomb's law analogous to $$G$$, constant in gravitational law.
$$m_{1}$$ and $$m_{2}$$ are mass analogous to $$q_{1}$$, $$q_{2}$$ charges.
A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are:
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constant, decreases, decreases
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increases, decreases, decreases
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constant, decreases, increases
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constant, increases, decreases
Explanation
Hint:
After charging, the capacitor is isolated. The capacitance of a capacitor is inversely proportional to its potential difference.
Explanation:
Step 1: Concept used:
Since the capacitor is isolated, the charge will remain constant.
Capacitance of parallel-plate capacitor is given by:
$$C = \cfrac{\varepsilon_o A}{d}$$ ..............(1)
Where, $$ \varepsilon_o $$ is the permittivity of free space.
Step 2: Conclusion:
From eq. (1)
$$C \ \alpha \ \dfrac {1}{d}$$
As $$d$$ is increased, capacitance will decrease.
By definition of capacitance,
$$Q = CV$$ (As $$Q$$ is constant)
$$V \ \alpha \ \dfrac {1}{C}$$
Since $$Q$$ is constant, $$V$$ will increase and
$$C$$ is decreased,
Hence option D is correct.
Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant K. The remaining half contains air as shown in the figure. The capacitor is now given a charge Q. Then
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An electric field in the dielectric-filled region is higher than that in the air-filled region
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On the two halves of the bottom plate the charge densities are unequal
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Charge on the half of the top plate above the air-filled part is $$\dfrac{Q}{K +}$$
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Capacitance of the capacitor shown above is $$(1+K)\dfrac{C_0}{2}$$, where $$C_0$$ is the capacitance of the same capacitor with the dielectric removed
Explanation
Parallel
Capacitance of the capacitor shown above is
$$C_eq = (1 + k)\cfrac{C_0}{2}$$
correct
Electric field in the dielectric-filled region is higher than that in the air-filled region
- incorrect as $$V\downarrow $$
Charge on the half of the top plate above the air filled;
$$Q_1 = \cfrac{C_0 Vk}{2}$$
And, with air filled;
$$Q_2 = \cfrac{C_0 V}{2}$$ Potential is same
$$\Rightarrow \cfrac{Q}{k}$$
Charge on the half of the top plate above the air-filled part is
- correct
A pendulum ( positively charged and hinged at some length above the plate) is swinging above a parallel plate (infinitely large and having negative charge),now consider the following statements, (consider gravity)
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angular momentum about the hinge point of the ball will be max at lowest point
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electric potential energy will be max at highest point
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gravitational potential energy will be lowest at highest point
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none of the above
Explanation
Angular momentum about the hinge point of the ball will be max at point. $$mV$$ will be max at closer point because law of conservation says that the swinging energy will be gathered when it will be at lowest point. The potential energy is maximum.
Which of the following is a vector quantity?
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Electric potential
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Electric feild
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Magnetic potential
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Magnetic feild
Explanation
Among the given physical quantities, electric and magnetic fields are vector quantities as they have both magnitude and direction.
Answer-(B),(D)
In Fig., if the potential at point B is taken as zero, then the potential at point A will be?
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$$8$$V
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$$16$$V
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$$24$$V
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None of the above
The SI unit of electric potential is
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$$Vm^{-1}$$
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$$C$$
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$$NC^{-1}$$
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$$V$$
Explanation
Electric potential is defined as the work done to move a unit positive charge from one point to another point.
The SI unit of electric potential is $$volt(V)$$.
Answer-(D)
Three charges $$-q,Q \ and \ -q$$ are placed at equal distances on a straight line. If the total potential energy of the system of three charges is zero, then the ratio $$Q:q$$ is :
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$$1:2$$
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$$2:1$$
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$$1:1$$
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$$1:4$$
Explanation
$$\textbf{Step 1 : Figure of arrangement}$$ $$\textbf{[Ref.Fig.]}$$
$$\textbf{Step 2 : Expression of potential Energy of system of charged particles.}$$
As potential energy is scalar quantity, so it will be added for all the combinations:
$$U=\dfrac{KQ_1Q_2}{d}+\dfrac{KQ_2Q_3}{d}+\dfrac{KQ_1Q_3}{2d}$$
$$\Rightarrow\ \ U=\dfrac{-KQq}{d}-\dfrac{KQq}{d}+\dfrac{Kq^2}{2d} = 0$$ (
Given that, the Total Potential Energy of system is equal to zero)
$$\Rightarrow\ \ q^2=4qQ$$
$$\Rightarrow\ \ Q:q=1:4$$
Hence ratio od $$Q:q$$ is equal to $$1:4$$
Option $$D$$ correct.
An electron of mass $$M_e$$, initially at rest, moves through a certain distance in a uniform electric field in time $$t_1$$. A proton of mass $$M_p$$ also intially at rest, takes time $$t_2$$ to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio $$t_2 / t_1$$ is nearly equal to :
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$$1$$
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$$\sqrt{\dfrac{M_p}{M_e}}$$
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$$\sqrt{\dfrac{M_e}{M_p}}$$
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$$1836$$
Explanation
The acceleration of electron under electric field $$E$$ is $$a_e=\dfrac{F}{M_e}$$
$$=\dfrac{qE}{M_e}$$
Similarly acceleration of proton under electric field=$$a_p=\dfrac{F}{M_p}$$
$$=\dfrac{qE}{M_p}$$
Since the distance traveled by both is same, we can write
$$s=\dfrac{1}{2}a_et_1^2=\dfrac{1}{2}a_pt_2^2$$
$$\implies \dfrac{t_2}{t_1}=\sqrt{\dfrac{a_e}{a_p}}$$
$$=\sqrt{\dfrac{M_p}{M_e}}$$
is equal to th
e work done by an external agent in carrying a unit of positive charge from the arbitrarily chosen reference point (usually infinity) to that point without any acceleration.
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Electric potential energy
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Electric field
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Electric potential
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Electric potential difference
Explanation
Electric potential is equal to work done by an external agent in carrying a unit positive charge from the arbitrary chosen reference point to that point without any acceleration.
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