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## Saturday, May 22, 2010

### AP Physics C - Multiple Choice Practice Questions on Electrostatics

Multiple choice as well as free response practice questions on electrostatics were discussed earlier on this blog. You can access them by clicking on the label ‘electrostatics’ below this post or by trying a search using the search box provided on this page. Today we will discuss a few more multiple choice practice questions in this section:

(1) Ten identical capacitors, each of value 2 μF are connected in series and this series combination is connected across a regulated power supply of output 10 V. The energy stored in any one of the 2 μF capacitors is

(a) 1 J

(b) 2 J

(c) 5 J

(d) 10 J

(e) 100 J

The effective capacitance of the series combination is 2 μF/10 = 0.2 μF.

[If n capacitors each of value C are in series, the effective capacitance is C/n].

The energy of the series combination on charging with the 10 volt supply is ½ CV2 = ½×0.2×102 = 10 J.

Therefore, the energy of one capacitor = 10 J/10 = 1 J.

(2) Two equal positive charges of value 8 μC are placed in a region of space where there are no external fields. When a third charge q is placed at the mid point of the line joining the other two charges, the system is found to be in equilibrium. The third charge q must be

(a) 8 μC

(b) – 8 μC

(c) 4 μC

(d) – 4 μC

(e) – 2 μC

The third charge q must be negative so that the mutual repulsive force between the positive charges (Q each, let us say) is counteracted by the attractive force exerted by the negative charge.

Of course the above fact will come out from the expression for the net force on Q, which we will put equal to zero. Therefore we have

(1/4πε0)Q2/r2 + (1/4πε0)Qq /(r/2)2 = 0

Or, Q = – 4q so that q = – Q/4 = – 8 μC/4 = – 2 μC (3) The figure shows 4 identical parallel metallic plates [(1), (2), (3) and (4)] arranged with equal separation d between neighbouring plates. The surface area of one side of each plate is A and the medium between the plates is air. Alternate plates are joined to terminals T1 and T2 so that the system makes a parallel plate capacitor. Suppose there are n plates (instead of 4) in the system where n > 1 and may be odd or even. What is the capacitance of the system made of these n plates?

(a) 2n ε0A/d

(b) n ε0A/d

(c) (n – 1)ε0A/d

(d) (n + 1)ε0A/d

(e) n ε0A/2d

The lower surface of plate (1) and the upper surface of plate (2) makes a capacitor of capacitance ε0A/d. The lower surface of plate (2) and the upper surface of plate (3) makes another capacitor of capacitance ε0A/d. Similarly the lower surface of plate (3) and the upper surface of plate (4) makes a third capacitor of capacitance ε0A/d. These three capacitors are connected in parallel and the system give a total capacitance of 3ε0A/d.

If there are n plates, the effective capacitance C will be given by

(4) In the combination of capacitors shown in the adjoining figure, what is the effective capacitance between the terminals A and B?

(a) 20 C

(b) 11 C

(c) 8 C

(d) 6 C

(e) 3 C

On connecting a voltage source between the terminals A and B, the potential at the junction of the two capacitors of value 2C is the same as the potential at the junction of the two capacitors of value 4C.

[The capacitors need not necessarily be equal. It is enough that the ratios of capacitance are equal to balance the Wheatstone bridge]

The capacitors of value 8C is therefore connected between equi-potential points and it can be ignored (since it does not get charged). The network thus reduces to four capacitors with the series combination of 2C and 2C connected in parallel with the series combination of 4C and 4C. Therefore, the effective value is C + 2C = 3C.

(5) Point charges +2q, +2q and –q are placed at the vertices A, B and C respectively of an equilateral triangle ABC of side 2a. How much external work is to be done to move these charges so that the side of the equilateral triangle becomes a?

(a) (1/4πε0)(4q2/a)

(b) – (1/4πε0)(4q2/a)

(c) (1/4πε0)(2q2/a)

(d) – (1/4πε0)(2q2/a)

(e) Zero

The external work W to be done is given by

W = U2 U1 where U2 and U1 are respectively the final and initial electrostatic potential energies of the system.

Now, U2 = (1/4πε0)[(2q×2q)/a + 2q(–q)/a + (–q)2q/a] = 0

U1 = (1/4πε0)[(2q×2q)/2a + 2q(–q)/2a + (–q)2q/2a] = 0

Therefore, W = 0. (6) A conducting sphere of radius R is arranged concentrically inside a thin conducting spherical shell of radius 2R. The sphere carries a charge +q and the spherical shell carries a charge –Q. The potential difference between the sphere and the shell is

(a) (1/4πε0)(q/2R)

(b) (1/4πε0)(q/R Q/2R)

(c) (1/4πε0)(Q/2R)

(d) (1/4πε0) (qQ)/2R

(e) (1/4πε0) (qQ)/R

The potential V1 of the inner sphere is equal to the sum of the potentials due to its own charge q and the charge –Q on the shell:

V1 = (1/4πε0)(q/R) +(1/4πε0)(–Q/2R)

[Note that the potential due to the shell is constant everywhere inside it and is equal to (1/4πε0)(–Q/2R)].

The potential V2 of the outer shell is equal to the sum of the potentials due to its own charge –Q and the charge q on the sphere inside it. Since the total charge contributing to the potential of the shell is q Q, its net potential is given by

V2 = (1/4πε0)(qQ)/2R

The potential difference between the sphere and the shell is given by

V1 V2 = (1/4πε0)[(q/R) –(Q/2R) –(q/2R) +(Q/2R)]

Or, V1 V2 = (1/4πε0) [(q/R) –(q/2R)] = (1/4πε0)(q/2R)

* * * * * * * * * * * * * * * *

You can easily work out this problem if you note that the potential due to the charge on the shell appears in the net potential of the sphere inside it. Therefore, when you find the potential difference between the sphere and the shell, the contribution by the shell gets canceled and it is enough to find the potentials due to the charge on the sphere (sphere alone) at its surface and at distance 2R and find the difference:

Therefore the answer is simply (1/4πε0) (q/R) – (1/4πε0) (q/2R), which is equal to (1/4πε0)(q/2R).

Now suppose we have a conducting sphere of radius R1 carrying a charge Q1 arranged concentrically inside a thin conducting spherical shell of radius R2 carrying a charge Q2. The potential difference between the sphere and the shell is(1/4πε0) (Q1/R1) – (1/4πε0) (Q1/R2), which is equal to (1/4πε0)[(Q1/R1) – (Q1/R2)].

1. 2. 