“Science distinguishes a Man of Honour from one of those Athletic Brutes whom
undeservedly we call Heroes.”

– John Dryden

Today we will discuss a few multiple choice
questions involving charging of capacitors. These questions will serve as
practice questions for various entrance examinations including the National
Eligibility Cum Entrance Test (NEET) for admission to MBBS and BDS Courses and
the Joint Entrance
Examination (JEE) for Admission to Undergraduate Engineering Programmes in
IITs, NITs and other Technical Institutions. Here are the questions:

(1) An initially uncharged capacitor of
capacitance 100 μF is charged at a steady rate of 20 μC/s. What is the time (in
seconds) required for the potential difference across the capacitor to build up
to 12 V?

(a) 240 s

(b) 120 s

(c) 60 s

(d) 30 s

The final charge

*Q*on the capacitor is given by*Q = CV =*100 μF×12 V = 1200 μC

Since the charging rate is 20 μC/s, the time

*t*required for the potential difference across the capacitor to build up to 12 V is given by*t*= 1200 μC/20 μC/s = 60 s

(2) A battery of emf

*V*volt is connected to a combination of four identical capacitors (each of capacitance*C*farad) arranged as shown in the adjoining figure. What is the amount of energy supplied by the battery to the capacitor combination?
(a)

*CV*^{2}/4
(b)

*CV*^{2}/2
(c)

*CV*^{2}/3
(d) 2

*CV*^{2}/3
The effective capacitance connected across the
battery is

*C +*(*C*/3) = 4*C*/3
The energy

*E*of the capacitor combination on getting charged to V volt is given by*E =*½ × (4

*C*/3)

*V*

^{2}= 2

*CV*

^{2}/3

(3) A parallel plate capacitor with air as
dielectric has capacitance

*C*. It is charged fully using a battery of emf*V*. The battery is then disconnected and the separation between the plates of the capacitor is doubled. What is the final energy stored in the capacitor?
(a) ½

*CV*^{2}
(b)

*CV*^{2}
(c) 3

*CV*^{2}/2
(d) 2

*CV*^{2}
(e)

*CV*^{2}/4
The charge

*Q*on the capacitor is*unchanged*since the battery is disconnected. The charge*Q*is given by*Q = CV*------------- (i)

When the separation between the plates is

*doubled*, the capacitance is*halved*(*C*/2).
[Note that the capacitance of a parallel plate
air capacitor is given by

*C =**ε*where_{0}A/d*A*is the plate area and*d*is the separation between the plates.
The energy

*E*stored in the capacitor is given by*E = Q*

^{2}/2

*C*

_{1}where

*C*

_{1 }=

*C/*2

Substituting for

*Q*from Eq. (i), we have*E = C*

^{2}

*V*

^{2}/(2

*C/*2) =

*CV*

^{2}, as given in option (b).

[Note that the increased energy of the capacitor
is the result of the external work done for increasing the separation between
the plates, overcoming the electrostatic attractive force between the plates]

(4) In the above question, suppose the battery
remains connected to the capacitor while doubling the separation. What will be
the final energy stored in the capacitor>

(a) ½

*CV*^{2}
(b)

*CV*^{2}
(c) 3

*CV*^{2}/2
(d) 2

*CV*^{2}
(e)

*CV*^{2}/4
In this case charges can flow through the wires
connecting the battery to the capacitor. The capacitor will hold a smaller
amount of charge (let us say,

*Q*_{1}) since the capacitance is decreased from*C*to*C*/2. We have\*Q*

_{1}= (

*C/*2)

*V*

The final energy (

*E*_{1}) stored in the capacitor is given by*E*

_{1}=

*Q*

_{1}

^{2}/2

*C*

_{1}= [(

*C/*2)

*V*]

^{2}/(2

*C/*2) =

*CV*

^{2}/4

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