Explanation
Out of the given statements which of the following are true :
A) Work done in moving a charge on equipotential surface is zero.
B) Electric lines of force are always normal to equipotential surface.
C) When two like charges are brought closer, the electrostatic potential energy of the system is decreased.
D) Electric lines of force converge at positive charge and diverge at negative charge.
When a dielectric material is introduced between the plates of a charged condenser, after disconnecting the battery, the electric field between the plates:
A condenser is charged and then battery is removed. A dielectric plate is put between the plates of condenser, then correct statement is
Statement(A): Negative charges always move from a higher potential to lower potential point
Statement (B): Electric potential is vector.
A parallel plate condenser is charged by connecting it to a battery. Without disconnecting the battery, the space between the plates is completely filled with a medium of dielectric constant k. Then:
Twenty seven identical mercury drops each charged to $$10V$$, are allowed to form a big drop. The potential of the big drop is
(1): The dielectric medium between the plates of a parallel plate capacitor lowers the potential difference between the plates without a battery.
(2): The maximum electric field that a dielectric can withstand without causing it to break down is dielectric strength.
Hint: apply gauss law for the two plates (conductors) where E=0.
Step 1: imagine a Gaussian surface
Imagine a Gaussian surface surrounding the two plates as shown in figure.
Step 2: Apply gauss law
Gauss law: electric flux through any closed surface is directly
proportional to the charge enclosed in it.
$$\oint{E.dA}$$ =$$\dfrac{q}{{{\varepsilon _o}}}$$
As both plates are conductor hence E=0 for the Gaussian surface
$$\dfrac{q}{{{\varepsilon _o}}}$$= 0
$${\varepsilon _o} \ne 0$$
So q=0
Step 3: find charge on each plate
As, total q=0
So charge on each plate will be equal and opposite
Final step: The charge on the smaller plate will be equal to other plate to make total charge zero on both the plates irrespective of shape and size.
A dielectric of thickness $$5$$ cm and a dielectric constant $$10$$ is introduced between the plates of a parallel plate capacitor having plate area $$500 sq.$$ cm and separation between the plates $$10cm$$. The capacitance of the capacitor with the dielectric slab is $$\varepsilon _{0}=8.8\times 10^{-12}C^{2}/N-m^{2}$$
Charges $$+q$$, $$-4q$$ and $$+2q$$ are arranged at the corners of an equilateral triangle of side $$0.15\ m$$. If $$q=1\ \mu C $$, their mutual potential energy is roughly:
Calculate the electrostatic potential energy of an electron-proton system of hydrogen atom. In the first Bohr orbit of hydrogen atom, the radius of the orbit is $$5.3\times 10^{-11}m\ $$:
A charge of $$2\ C$$ is moved from a point $$2\ m$$ away from a charge of $$1\ C$$ to a point $$1\ m$$ away from that charge. The work done is:
A positive point charge q is carried from a point B to a point A in the electric field of a point charge +Q. If the permittivity of free space is $$\epsilon _{0}$$ , the work done in the process is given by
A, B, C are three points on a circle of radius $$1 cm$$. These points form the corners of an equilateral triangle. A charge $$2C$$ is placed at the centre of the circle. The work done in carrying a charge of $$0.1\mu C$$ from A to B is :
The capacitance of a capacitor becomes $$\dfrac{7}{6}$$ times its original value if a dielectric slab of thickness, $$t=\dfrac{2}{3}d$$ is introduced in between the plates. d is the separation between the plates. The dielectric constant of the dielectric slab is :
The potential difference across $$3 \mu F$$ condenser is :
The equivalent capacitance of three capacitors of capacitance $$C_{1}, C_{2}$$ and $$C_{3}$$ connected in parallel is 12 units and the product $$C_{1}C_{2}C_{3}=48$$. When the capacitors $$C_{1}$$ and $$C_{2}$$ are connected in parallel the equivalent capacitance is 6 units. Then the capacitance are :
Two condensers of capacitance $$4\mu F$$ and $$5\mu F$$ are joined in series. If the potential difference across $$5\mu F$$ is $$10V$$, then the potential difference across $$4\mu F$$ condenser is :
In the capacitor of capacitance $$20\ \mu F$$, the distance between plates is $$2\ mm$$. If a material of dielectric constant $$ 2$$ is inserted between the plates, then the capacitance of the system is :
Electric force $$F=qE$$
given, $$F=3000 N, q=3 C$$
$$\therefore E=\dfrac{F}{q}=\dfrac{3000}{3}=1000 V/m$$
Potential difference, $$V={E}{d}$$ where $$d=0.01 m$$
$$\therefore V=1000\times 0.01=10 V$$
Ans:(A)
Two metal plates are separated by a distance $$d$$ in a parallel plate condenser. A metal plate of thickness $$t$$ and of the same area is inserted between the condenser plates. The value of capacitance increases by a factor of :
Hint:
Capacitance of a parallel plate capacitor with distance $$d$$ between the plates is given by,
$$C=\dfrac{A{{E}_{o}}k}{d}$$
Where,
$$A$$ is area of plates,
$${{E}_{o}}$$ is permittivity of free space,
$$k$$ is dielectric constant of medium filled between plates.
Capacitance of a parallel plate capacitor with distance $$d$$ between the plates, with a dielectric slab of thickness $$t$$ and dielectric constant$$k$$ between the plates is given by,
$$C=\dfrac{A{{E}_{o}}}{d-t\left( 1-\dfrac{1}{k} \right)}$$
Note that for metal, value of $$k$$ is infinite.
Step 1: Given.
A parallel plate capacitor with distance d between plates of area A. Capacitance is given by,
$$C=\dfrac{A{{E}_{o}}}{d}$$
Step 2: Calculate new capacitance in terms of old capacitance.
A metal plate of thickness t is inserted between them. New capacitance is:
$$C'=\dfrac{A{{E}_{o}}}{d-t\left( 1-\dfrac{1}{k} \right)}$$
$$\Rightarrow C'=\dfrac{A{{E}_{o}}}{d-t}$$ (Since, k is infinite for metal)
$$\Rightarrow C'=\dfrac{A{{E}_{o}}}{d(1-\dfrac{t}{d})}$$
$$\Rightarrow C'=\dfrac{C}{(1-\dfrac{t}{d})}$$
$$\Rightarrow C'=\dfrac{1}{(1-\dfrac{t}{d})}C$$ Thus value of capacitance increases by a factor $$\dfrac{1}{(1-\dfrac{t}{d})}$$
Option D is correct.
Three capacitors $$2\mu F, 3\mu F$$ and $$5\mu F$$ are connected in parallel. The capacitance of the combination:
The equivalent capacity between the points X and Y in the circuit with $$C=1\ \mu F$$ is:
A capacitor of $$10 \mu F$$ capacitance is charged by a $$12 V$$ battery. Now the space between the plates of capacitors is filled with a dielectric of dielectric constant $$K = 3$$ and again it is charged. The magnitude of the charge is :
Three capacitors $$2\mu F, 3\mu F$$ and $$6\mu F$$ are connected in series. The effective capacitance of the combination is:
A neutral hydrogen molecule has two protons and two electrons. If one of the electrons is removed we get a hydrogen molecular ion $$\left ( H_{2}^{+} \right )$$ . In the ground state of $$ H_{2}^{+} $$ the two protons are separated by roughly $$1.5$$$$\ A^o$$ and the electrons is roughly $$1$$ $$A^o$$ from each proton. The potential energy of the system is :
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