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## Sunday, March 13, 2011

### AP Physics C – Answer to Free Response Practice Question on Electrostatic Potential Energy

“Genius is one per cent inspiration and ninety nine per cent perspiration.”

Thomas A. Edison

A free response practice question on Electrostatic Potential Energy was given to you in the post dated 11th March 2011. As promised, I give below a model answer along with the question: A thin circular ring of radius R has positive charges sprayed uniformly along it so that the linear charge density along it is λ. The ring is located in the y-z plane with its centre at the origin of a right handed Cartesian co-ordinate system, in a space lab, where the effect of gravity is negligible. From the point P [–(√15)R, 0, 0] shown in the figure, a particle of mass m and charge +q is projected with speed v along the positive x-direction. Now answer the following:

(a) Which one of the following statements (i), (ii) and (iii) regarding the electric field vector at P (due to the charges on the ring) is correct? [Put a tick (√) mark against your choice].

(i) Electric field at P is along the positive x-direction ……..

(ii) Electric field at P is along the negative x-direction ……..

(iii) Electric field at P is zero ……..

(b) Determine the electrostatic potential energy of the charge q when it is located at P.

(c) Determine the electrostatic potential energy of the charge q when it is located.at the centre of the ring.

(d) Determine the minimum value of the speed of projection for the charged particle to pass through the ring.

(e) Explain what happens to the charged particle if the speed of projection is less than the minimum value mentioned in part (d) above.

(a) (ii) Electric field at P is along the negative x-direction …√…..

Since the point P is on the axis of the charged ring, the electric field at P is directed along the axis. Since the charge on the ring is positive, the electric field is directed from the ring to the point P. So the field is along the negative x-direction.

The direction of the field due to the charge at a point A on the ring is directed along the line AP. This field can be resolved into rectangular components, one along the axis of the ring and the other normal to the axis. When we consider the electric field due to an equal charge situated at the diametrically opposite point B on the ring, we find that the normal components of the fields due to A and B are equal and opposite so that they get canceled. But the axial components are in the same direction (negative x-direction) and they get added up. This is why the field at P is along the negative x-direction.

(b) The total charge Q on the ring is given by

Q = 2πR λ.

The distance r of the point P from the charges is given by

r = [R2 + 15 R2]1/2 = 4R

Therefore, the electric potential V1 at the point P is given by

V1 = (1/4πε0)(Q/r) where ε0 is the permittivity of free space.

Substituting for Q and r, we have

V1 = (1/4πε0)( 2πRλ /4R) = λ/8ε0

The electrostatic potential energy E1 of the charge q when it is located at P is given by

E1 = V1q = λq/8ε0

(c) The electric potential V2 at the centre of the ring is given by

V2 = (1/4πε0)(Q/R) = (1/4πε0)( 2πRλ /R) = λ/8ε0 = λ/2ε0

The electrostatic potential energy E2 of the charge q when it is located at the centre of the ring is given by

E1 = V2q = λq/2ε0

(d) When the particle of mass m and charge +q is projected (along the axis) towards the ring, its potential energy increases until it reaches the centre of the ring. The increase in potential energy is at the cost of the kinetic energy of the particle. Once the particle just passes the centre of the ring, it can continue to move forward and gain kinetic energy at the cost of its potential energy.

If the initial kinetic energy with which the particle is projected from the point P is equal to the difference between its electrostatic potential energies at the points P and the centre of the ring, the particle will reach the centre of the ring with zero speed. Therefore, the minimum kinetic energy with which the charged particle is to be projected so as to pass through the ring is equal to the difference between its electrostatic potential energies at the points P and the centre of the ring. Therefore, we have

½ mv2 = E1 E2 where v is the minimum speed with which the charged particle is to be projected.

Substituting for E1 and E2 we have

½ mv2 = (λq/2ε0) – (λq/8ε0) = 3λq/8ε0

This gives v = (3λq/4ε0m)1/2

(e) If the speed of projection of the charged particle is less than the minimum value mentioned in part (d) above, it will move towards the ring with its speed decreasing gradually. Its kinetic energy will become zero before reaching the centre of the ring and it will momentarily come to rest. Under the action of the electric field directed along the negative x-direction, the particle will turn back and will retrace its path and will move past the point P all the way to infinite distance from the ring before coming to rest.

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