An isolated metal sphere is charged to a potential V. The charge on the sphere is q. The charge on the sphere may be considered to act as a point charge at the centre of the sphere.

Question 2

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

(b) An isolated metal sphere is charged to a potential V. The charge on the sphere is q.

The charge on the sphere may be considered to act as a point charge at the centre of the sphere.

The variation with potential V of the charge q on the sphere is shown in Fig. 5.1.

Fig. 5.1

Use Fig. 5.1 to determine

(i) the radius of the sphere,                                                                 [2]

(ii) the energy required to increase the potential of the sphere from zero to 24 kV. [3]

(c) The sphere in (b) discharges by causing sparks when the electric field strength at the surface of the sphere is greater than 2.0 × 10⁶ V m^-1.

Use your answer in (b)(i) to calculate the maximum potential to which the sphere can be

charged. [3]

Reference: Past Exam Paper – November 2014 Paper 43 Q5

Solution:

(a) The electric potential at a point is defined as the work done / energy in moving unit positive charge from infinity (to the point)

(b) (i)

V = q / 4πε₀r

{from the graph,} at 16 kV, q = 3.0 × 10^–8 C

{r = q / 4πε₀V}

r = (3.0 × 10^–8) / (4π × 8.85 × 10^–12 × 16 × 10³)                    C1

r = 1.69 × 10^–2 m        (allow 2 s.f.)                                        A1

(ii)

The energy is represented by the area ‘below’ line

{Energy = area of triangle between V = 0 kV to V = 24 kV}

{when V = 24 kV, q = 4.5 × 10^–8 C}

Energy = ½ qV

Energy = ½ × 24 × 10³ × 4.5 × 10^–8

Energy = 5.4 × 10^–4 J                        

(c)

since V = q / 4πε₀r      and      E = q / 4πε₀r²              giving Er = V   (or V = E/r)

{The maximum electric fields strength at which the sphere can be charged is E = 2.0 × 10⁶ V m^-1.}

2.0 × 10⁶ × 1.7 × 10^–2 = V

V = 3.4 × 10⁴ V