Two positively charged metal spheres A and B are situated in a vacuum, as shown in Fig. 5.1.

Question 23

(a) Define electric potential at a point. [2]

(b) Two positively charged metal spheres A and B are situated in a vacuum, as shown in Fig. 5.1.

Fig. 5.1

A point P lies on the line joining the centres of the two spheres and is a distance x from the surface of sphere A.

The variation with x of the electric potential V due to the two charged spheres is shown in Fig. 5.2.

Fig. 5.2

(i) State how the magnitude of the electric field strength at any point P may be determined from the graph of Fig. 5.2. [1]

(ii) Without any calculation, describe the force acting on a positively charged particle placed at point P for values of x from x = 0 to x = 10 cm. [3]

(c) The positively charged particle in (b)(ii) has charge q and mass m given by the expression

q / m = 4.8 × 10⁷ C kg^-1.

Initially, the particle is at rest on the surface of sphere A where x = 0. It then moves freely along the line joining the centres of the spheres until it reaches the surface of sphere B.

(i) On Fig. 5.2, mark with the letter M the point where the charged particle has its maximum speed. [1]

(ii) 1. Use Fig. 5.2 to determine the potential difference between the spheres. [1]

2. Use your answer in (ii) part 1 to calculate the speed of the particle as it reaches the surface of sphere B.

Explain your working. [3]

Reference: Past Exam Paper – June 2015 Paper 42 Q5

Solution:

(a) Electric potential at a point is defined as the work done per unit positive charge in bringing the positive charge from infinity to the point.

(b)

(i) The electric field strength is given by the gradient of the V-x graph.

(ii)

maximum at surface of sphere A       or at x = 0 (cm)

zero at x = 6 (cm)

then increases but in opposite direction

{Electric field strength E = (-) potential gradient

Force = Eq = gradient × q

The force is maximum at the surface of A (x = 0 cm) (as the gradient has the largest value).

The force becomes zero at x = 6 cm (as gradient = 0).

The force then increases but in the opposite direction (as gradient increases but now has a different sign).}

(c)

(i) M shown between x = 5.5 cm and x = 6.5 cm

{Gain in KE = Loss in electric potential energy

Gain in KE = ΔV × q

The gain in KE (and thus, speed) is maximum when the change in potential ΔV has the largest value.

Largest ΔV = Vmax (at surface of A) – Vmin (at x = 6.0 cm)}

(ii) 1. ΔV = (570 – 230) = 340 V         (allow 330 V to 340 V)

2.

{Considering the energy changes,}

q(Δ)V = ½mv²                        or change/loss in PE = change/gain in KE     or ΔE_K = ΔE_P

{q/m × ΔV = ½ v²

q/m = 4.8 × 10⁷ C kg^-1 }

4.8 × 10⁷ × 340 = ½ v²

v² = 3.26 × 10¹⁰

v = 1.8 × 10⁵ m s^–1