# Questions OBJECTIVE - II and Answer Capacitors HC Verma Part II

Q#1

The capacitance of a capacitor does not depend on

(a) the shape of the plates

(b) the size of the plates

(c) the charges on the plates

(d) the separation between the plates.

The capacitance of a capacitor depends on the shape of the plates. It is different for parallel-plate, spherical or cylindrical capacitors.

It also depends on the size of the plates ie the area of the plates. With the increase in the area the capacitance increases.

The capacitance depends on the separation of the plates. It decreases with the increase in the separation.

So, the options (a), (b), and (d) are incorrect.

The capacitance does not depend on the charges on the plates. In fact, it is the ratio of Q and V and a proportionality constant.

Hence only option (c) is correct.

Q#2

A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

(a) the electric field in the capacitor

(b) the charge on the capacitor

(c) the potential difference between the plates

(d) the stored energy in the capacitor.

The electric field does not remain the same because an opposite electric field developes inside the dielectric slab. And the net electric field is different.

Potential difference between the plates, V =Ed, where E is the net electric field inside the capacitor. Since this net electric field does not remain the same, V also changes.

Energy stored per unit volume of a capacitor, u = $\frac{1}{2}\epsilon_0E^2$. Since E is variable so the energy sored will not remain the same.

Thus the options (a), (c) and (d) are incorrect.

Since the capacitor is isolated hence the charge on it will remain unchanged.

Option (b) is correct.

Q#3

A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is Q and the magnitude of the induced charge on each surface of the dielectric is Q'.

(a) Q' may be larger than Q,

(b) Q' must be larger than Q,

(c) Q' must be equal to Q,

(d) Q' must be smaller than Q.

The induced charge is given as,

Q' =Q(1 - $\frac{1}{K}$), where Q is the charge on the capacitor.

Since the dielectric constant K is positive, the factor (1 - $\frac{1}{K}$) will be less than 1.

Hence Q' < Q. So Q' must be less than Q. Option (d) is correct.

Q#4

Each plate of a parallel plate capacitor has a charge q on it. The capacitor is now connected to a battery. Now,

(a) the facing surfaces of the capacitor have equal and opposite charges.

(b) the two plates of the capacitor have equal and opposite charges

(c) the battery supplies equal and opposite charges to the two plates.

(d) the outer surfaces of the plates have equal charges.

Total charge in a battery always remains zero. It supplies the equal and opposite charges to the plates of a capacitor. Option (c) is correct.

When the capacitor is connected to a battery the potential difference across the plate V is equal to the emf of the battery. The charge supplied due to this P.D. is, Q = CV. Initially, the same charge q is on each plate. Due to the same nature, they will repel each other and come to the outer surfaces of the plates. The electric field between the plates and inside the plates will be zero. After connection to the battery, one plate will have a surplus charge CV that will come to the inner surface of the plate to attract the -CV charge on the other plate. So finally, facing surfaces of the capacitor have equal and opposite charges CV. Option (a) is correct.

The previous charges q will be on the outer surfaces as in the beginning. Option (d) is correct.

Since the plates have charges q + CV and q - CV, option (c) is incorrect.

Q#5

The separation between the plates of a charged parallel plate capacitor is increased. Which of the following quantities will change?

(a) charge on the capacitor

(b) the potential difference across the capacitor

(c) the energy of the capacitor

(d) energy density between the plates.

The charge given to an isolated capacitor remains constant. When separation between the plates d is increased, the elctric field between the plates still remais the same,

E = $\frac{Q}{2A\epsilon_0}$

Now the energy density between the plates = $\frac{1}{2}\epsilon_0E^2$

Potential difference between the plates, V = Ed.

Here E is constant but d varies. So V changes.  Also, the energy of a capacitor U = $\frac{1}{2}QV$. Here Q is constant but V changes. So the energy of a capacitor will alaso change.

Thus the options (b) and (c) will change.

Q#6

A parallel plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor.

(a) The battery will supply more charge

(b) The capacitance will increase

(c) The potential difference between the plates will increase

(d) Equal and opposite charges will appear on the two faces of the metal plate.

Since the sheet is made of metal, equal and opposite charges will appear on both faces of the sheet under the influence of the charges on the plates of capacitor. Option (d) is correct.

So it is now three capacitors connected in series. With separations, x, t and d-x-t. Where x is distance of metal sheet from a plate and t is the thickness of the sheet. Their capacitances are,

C = $\frac{\epsilon_0 A}{x}$

C' = $\frac{\epsilon_0 A}{t}$

C" = $\frac{\epsilon_0}{A(d-x-t)}$

Equivalant capacitance C* is given as

$\frac{1}{C*}=\frac{1}{C}+\frac{1}{C'}+\frac{1}{C''}$

$\frac{1}{C*}=\frac{x}{\epsilon_0 A}+\frac{t}{\epsilon_0 A}+\frac{d-x-t}{\epsilon_0 A}$

$\frac{1}{C*}=\frac{x+t+d-x-t}{\epsilon_0 A}$

$\frac{1}{C*}=\frac{d}{\epsilon_0 A}$

$C*=\frac{\epsilon_0 A}{d}$

= Capacitance of the capacitor without the sheet.

So, with insertion of the thin metal sheet the capacitance is not increasing. Option (b) is incorrect.

Since capacitance is not changing, Q = CV, V is not changing because it is connected to the battery. Hence the charge Q supplied by the battery will not change. Option (a) and (c) are also incorrect.

Q#7

Following operations can be performed on a capacitor.

X - connect the capacitor to a battery of emf E.

Y - disconnect the battery.

Z - reconnect the battery with polarity reversed.

W - insert a dielectric slab in the capacitor.

(a) In XYZ (perform X, then Y, then Z) the stored electric energy remains unchanged and no thermal energy is developed.

(b) The charge appearing on the capacitor is greater after the action XWY than after the action XYW.

(c) The electric energy stored in the capacitor is greater after the action WXY than after the action XYW.

(d) The electric field in the capacitor after the action XW is the same as that after WX.

With the reconnection by reversed polarity, electric field will change. Hence the stored electric energy will change. Option (a) is incorrect.

Potential difference between the plates with a dielectric,

V = $\frac{Qd}{\epsilon_0 AK}$

When the plates are connected with a battery, V is constant. So,

Q = $\frac{V\epsilon_0 AK}{d}$      (*)

In the action XYW, Q is constant.

Here, V = $\frac{Qd}{\epsilon_0 A}$

Q = $\frac{V\epsilon_0 A}{d}$        (**)

Since K > 1, (i) is greater than (ii). So option (b) is true.

Capacitance with a dielectric, C = KCₒ

Energy stored,

E = $\frac{1}{2}CV^2$

E = $\frac{1}{2}C_0V^2$             (***)

In XYW, Q is constant and  = CₒV

Energy stored,

E' = $\frac{1}{2}QV$

E' = $\frac{1}{2}C_0V^2$                                         (****)

So, E = KE'

Since K > 1, E > E'.

Option (c) is correct.

In either of actions XW or WX the electric field in the capacitor,

E = $\frac{E_0}{K}$, where Eₒ is the electric field without the dielectric.

So it is same in both the cases. Option (d) is correct.