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## Tuesday, July 8, 2008

### AP Physics B and C- Electrostatics: Answers to Free Response Practice Questions on Electric Field and Potential

The world is a dangerous place, not because of those who do evil, but because of those who look on and do nothing.

- Albert Einstein

In the post dated 5th July 2008, two free response practice questions on electric field and potential were given to you. As promised, I give below model answers along with the questions:

(1) Two identical thin rings of radius R are arranged coaxially with a separation R. A charge +Q is sprayed uniformly on one ring while a charge –Q is sprayed uniformly on the other.

(a) Calculate the electric potentials at the centres of the rings

(b) What is the electric potential on the common axis, midway between the rings? Justify your answer.

(c) A point charge +q is moved very slowly from the centre of the positively charged ring to the cenre of the negatively charged ring. Calculate the work done by the external agency for moving this charge.

(d) The charge +q and the negatively charged ring are now moved to a very large distance from the positively charged ring. Now, determine the electric field at the cntre of the positively charged ring.

(a) The electric potential (V1) at the centre of the positively charged ring is the sum of the positive potential due to its own positive charge and the negative potential due to the negative charge on the other ring. Thus, putting k = 1/4πε0 where ε0 is the permittivity of free space (or air) , we have

V1 = k[(Q1/R) + (Q2/√2 R)] = k[(Q1/R) – (Q2/√2 R)]

[The distance of the charges on the negatively charged ring from the centre of the positively charged ring is R√2].

Similarly, the electric potential (V2) at the centre of the negatively charged ring is

V2 = k[(Q2/R) + (Q1/√2 R)] = k[(Q1/√2 R) – (Q2/R)]

(b) The electric potential on the common axis, midway between the rings is zero since the positive and negative charges on the rings produce positive and negative potentials of equal magnitude.

(c) The work (W) done in moving the positive charge +q from the centre of one ring to the centre of the other is the product (V1 V2)q.

Therefore, W = k[{(Q1/R) –(Q2/√2 R)} {(Q1/√2 R) – (Q2/R)}]q

Or, W = kq(Q1 + Q2) [(1/R) – (1//√2 R)] = (kq/√2 R) (Q1 + Q2) (√2 – 1)

The work done by the external agency moving the charge is negative since the positive charge will move by itself from the positive potential point to the negative potential point. So, the answer is W = (kq/√2 R) (Q1 + Q2) (1 √2)

(d) The electric field at the centre of the isolated positively charged ring is zero since a test charge placed at the centre of the ring will feel equal radial forces all around.

(2) Two small identical conducting spheres of radius R carrying charges Q1 and Q2 (Q1 > Q2) attract each other with a force of magnitude F when they are separated by a distance r (between their centres) in air. When the spheres are brought into contact and then separated to the initial distance r, they repel each other with a force of the same magnitude F. Now, answer the following questions:

(a) It is given that the charge Q1 is positive. Calculate the electric potential difference between the centres of the spheres before bringing them into contact.

(b) What are the charges on the two spheres after bringing them into contact? Justify your answer.

(c) Calculate the electric field midway between the spheres after bringing the spheres into contact and separating them to the distance r.

(d) Calculate the ratio of the initial charges (Q1/Q2) on the spheres.

(e) Keeping the separation at r itself, the charges on the spheres are now changed so that one sphere carries positive charge +4q and the other sphere carries negative charge –q. Calculate the distance of the null point (where the electric field is zero) from the centre of the negatively charged sphere.

(a) The electric potential (V1) at the centre of the positively charged sphere is the sum of the positive potential due to its own positive charge and the negative potential due to the negative charge on the other sphere. But the potential inside a conductor is the same as the potential on its surface (since there cannot be a field within a conductor). Therefore, the potential at the centre of the positively charged sphere carrying charge Q1 is given by

V1 = k[(Q1/R) Q2/(r R)] where k = 1/4πε0.

[The sign of Q2 is negative as indicated by the attractive force between the spheres. The distance of the surface of the positively charged sphere from the centre of the negatively charged sphere is r R and hence it appears in the above expression for potential].

The electric potential (V2) at the centre of the negatively charged sphere is given by

V2 = k[(Q2/R) + Q1/(r R)]

Potential difference between the cenres of the spheres is

V1 V2 = k[(Q1/R) Q2/(r R) {(Q2/R) + Q1/(r R)}]

= k[Q1(r R R) + Q2(r R R)] /[R(r R)]

= k (r 2R) (Q1 + Q2) / [R(r R)]

(b) When the spheres are brought into contact, they share the total charge equally since the spheres are identical. Since the total charge is Q1Q2, the charge on each sphere will be (Q1Q2)/2.

(c) The electric field midway between the spheres is zero since the spheres carry like charges (positive here) of the same magnitude so that a test charge placed at the point will experience equal and opposite forces.

(d) Since the initial attractive force and the final repulsive force between the spheres have the same magnitude F, we have, considering only the magnitudes of the charges,

F = kQ1Q2/r2 = k[(Q1Q2)/2]2/r2 where k =1/4πε0.

Therefore, Q1Q2 = (Q1Q2)2/4.

Or, 4Q1Q2 = Q12+ Q22 – 2 Q1Q2.

This can be written as Q1Q2[(Q1/ Q2) + (Q2/ Q1) – 6] = 0

Since Q1Q2 is not equal to zero, we have

(Q1/ Q2) + (Q2/ Q1) – 6 = 0

Putting Q1/ Q2 = x we have x + 1/x – 6 = 0

Or, x2 – 6x + 1 = 0

This gives x = 3 ± √8

The ratio of the magnitudes of the charges (Q1/ Q2) is thus 3 ± √8. Since Q2 is negative, the answer is – (3 ± √8).

So, two values are possible for the ratio of charges.

(e) The null point is nearer the smaller charge, but outside on the straight line passing through the centres of the spheres. The null point is produced because of the equal and opposite forces produced (on a test charge) because of the charges +4q and q. If the distance of the null point from the centre of the negatively charged sphere is d, the distance from the centre of the positively charged sphere is r+d. Since the net electric field at the null point is zero, we have

k×4q/(r+d)2 + k(– q) /d2 = 0

Or, 2/(r+d) = 1/d, which gives d = r.