Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 46

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Question 1
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Some charge is being given to a conductor. Then its potential:

A
is maximum somewhere between surface and centre.

B
is maximum at surface.

C
is maximum at centre.

D
remains same throughout the conductor.

Solution

Given that some charge is given to a conductor then the whole charge is distributed over its surface only. Inside of conductor, electric field is zero whereas potential is same as on the surface. Hence, throughout the conductor, potential is same i.e, the whole conductor is equipotential.
Question 2
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Point A is at a lower electrical potential than point B. An electron between them on the line joining them will

A
move towards A

B
move towards B

C
move at right angles to the line joining A and B

D
remain at rest

Solution

Given that point $$A$$ is at lower electric potential than point $$B$$. The electron between them on line joining will move.
We have to find where this electron moves.
Since we know that electric currents move from a higher potential or a lower potential. Also, electrons move in the direction opposite to electric current. So the electron on the line joining two points $$A$$ and $$B$$ will move from lower to higher potential i.e, it will move towards $$B$$.
Question 3
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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)

A
angular momentum about the hinge point of the ball will be max at lowest point

B
electric potential energy will be max at highest point

C
gravitational potential energy will be lowest at highest point

D
none of the above

Solution

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.
Question 4
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In context of a capacitor , which of the following statement is correct?

A
Electric field may be uniform

B
There must be some equipotential surface somewhere inside

C
Electric potential must be max at surface of one of the plate in case of parallel plate capacitor

D
None of the above

Solution

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.
Question 5
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A charge $$q$$ is placed at $$A(2,3,3)$$ in the $$XYZ$$ co-ordinate system. Find the electric potential at $$B(-2,3,3)$$

A
$$\dfrac{q}{4 \pi \epsilon_o}$$

B
$$\dfrac{q}{16 \pi \epsilon_o}$$

C
$$\dfrac{3q}{16 \pi \epsilon_o}$$

D
None of these

Solution

distance=$$r=\sqrt{2+2)^2+(3-3)^2+(3-3)^2}=4$$

$$V=\dfrac{q}{4\pi\epsilon_o r}$$

$$\implies V=\dfrac{q}{16\pi\epsilon_o}$$

Answer-(B)
Question 6
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The capacitance of a parallel plate capacitor with air as dielectric is C. If a slab of dielectric constant K and of the same thickness as the separation between the plates is introduced so as to fill 1/4th of the capacitor (shown in figure), then the new capacitance is

A
$$(K+2)\dfrac{C}{4}$$

B
$$(K+3)\dfrac{C}{4}$$

C
$$(K+1)\dfrac{C}{4}$$

D
None of these

Solution

Given,
Capacitance of a parallel plate capacitor with air as dielectric is $$=C$$
Dielectric constant of the slab$$=K$$
Sepration between the plates is $$=\dfrac{1}{4}^{th}$$
Capacitance, $$C=\dfrac{{\epsilon}_0A}{d}$$
As one-fourth of capacitor is filled with dielectric of constant K, then,
$$C_1=\dfrac{K{\epsilon}_0A/4}{d}$$
and $$C_2=\dfrac{{\epsilon}3A/4}{d}$$
Both $$C_1$$ and $$C_2$$ are in parallel.
$$\therefore C_p=C_1+C_2=\dfrac{K{\epsilon}_0A}{4d}+\dfrac{3{\epsilon}_0A}{4d}$$
$$=(K+3)\dfrac{{\epsilon}_0A}{4d}=(K+3)\dfrac{C}{4}$$
So, The new capacitance $$=(K+3)\dfrac{C}{4}$$.
Question 7
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Two point charges, each of charge $$q$$, are placed at a separation of $$2a$$. The electric potential at their midpoint will be :

A
Zero

B
$$\dfrac{q}{2 \pi {\varepsilon}_{0} a}$$

C
$$\dfrac{q}{8 \pi {\varepsilon}_{0} a}$$

D
$$\dfrac{q}{2 \pi {\varepsilon}_{0} {a}^{2}}$$

Solution

Potential at any point is given by $$V = \dfrac{q}{4\pi \epsilon_o d}$$
where $$d$$ is the distance of that point from the charge.
Thus potential at the midpoint P $$V_p = \dfrac{q}{4\pi \epsilon_o a} + \dfrac{q}{4\pi \epsilon_o a}$$
$$\therefore$$ $$V_p = \dfrac{q}{2\pi \epsilon_o a}$$
Question 8
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In given circuit, $$C_{1}=C_{2}=C_{3}=C$$ initially. Now, a dielectric slab of dielectric constant $$K=\frac{3}{2}$$ is inserted in $$C_{2}$$.
The equivalent capacitance become

A
$$\dfrac{5C}{7}$$

B
$$\dfrac{7C}{5}$$

C
$$\dfrac{2C}{3}$$

D
$$\dfrac{C}{2}$$

Solution

When a dielectric slab of dielectric constant $$\displaystyle K=\cfrac{3}{2}$$ is inserted between the plates of $$C_{2}$$, its new capacitance $$\left ( {C_{2}}' \right )$$ becomes
$$\displaystyle \left ( {C_{2}}' \right )=\cfrac{3}{2}C$$
Equivalent capacitance of $$\left ( {C_{2}}' \right )$$ and $$C_{3}$$ is
$$\displaystyle C_{eq}={C_{2}}'+C_{3}=\cfrac{3}{2}C+C=\cfrac{5C}{2}$$
Now, $$C_{eq}$$ and $$C_{1}$$ are in series. Therefore, their equivalent capacitance is
$$\displaystyle C_{eq}=\cfrac{C_{eq}\times C_{1}}{C_{eq}+ C_{1}}=\cfrac{\cfrac{5C}{2}\times C}{\cfrac{5C}{2}+ C}$$
$$\displaystyle =\cfrac{5C^{2}}{7C}=\cfrac{5C}{7}$$
Question 9
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The potential difference between the two plates of a parallel plate capacitor is constant. When air between the plates is replaced by dielectric material, the electric field intensity :

A
Decreases

B
Remains unchanged

C
Becomes zero

D
Increases

Solution

In general capacitance of parallel plate capacitor is given by:
$$C=\dfrac{k\epsilon_0 A}{d}$$
where C is capacitance, k is relative permittivity of dielectric material, $$\epsilon_0$$ is permittivity of free space constant, A is area of plates and d is distance between them.

Therefore, the capacitance of parallel plates is increased by the insertion of a dielectric material. Further, the capacitance is inversely proportional to the electric field between the plates, and hence the presence of the dielectric decreases the effective electric field.
Question 10
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The difference in the effective capacity of two similar capacitors when joined in series and then in parallel is $$6 \mu F$$. The capacity of each capacitor is

A
$$2 \mu F$$

B
$$4 \mu F$$

C
$$8 \mu F$$

D
$$16 \mu F$$

Solution

Let the capacitance of each capacitor be $$C$$.
Equivalent capacitance of the two capacitors connected in parallel $$C_p = C+C = 2C$$
Equivalent capacitance of the two capacitors connected in series $$C_s = \dfrac{CC}{C+C} = \dfrac{C}{2}$$
But, $$C_p - C_s = 6\mu F$$
Or $$2C - \dfrac{C}{2} = 6\mu F$$
OR $$\dfrac{3C}{2} = 6\mu F$$
$$\implies$$ $$C = 4\mu F$$