A solid metal sphere of radius R is isolated in space. The sphere is positively charged so that the electric potential at its surface is VS.

Question 26

A solid metal sphere of radius R is isolated in space. The sphere is positively charged so that the electric potential at its surface is VS. The electric field strength at the surface is ES.

(a) On the axes of Fig. 6.1, show the variation of the electric potential with distance x from the centre of the sphere for values of x from x = 0 to x = 3R.

Fig. 6.1

[3]

(b) On the axes of Fig. 6.2, show the variation of the electric field strength with distance x from the centre of the sphere for values of x from x = 0 to x = 3R.

Fig. 6.2

[3]

[Total: 6]

Reference: Past Exam Paper – June 2012 Paper 41 & 43 Q6

Solution:

(a)

sketch: from x = 0 to x = R, potential is constant at VS

smooth curve through (R, VS) and (2R, 0.5VS)       

smooth curve continues to (3R, 0.33VS)

{Electric potential: V = Q / 4πϵ₀x        where x is the distance from the centre

For a conductor, the potential is constant inside the sphere. So, from x = 0 to x = R (that is, inside the sphere), potential = V_S and is constant. This is a horizontal line in the graph.

Hence, we also have the point (R, V_S).

V_S = Q / 4πϵ₀R

From the equation for potential above, the potential V is inversely proportional to x. The greater the value of x, the smaller the potential.

V = Q / 4πϵ₀x

When x = R, V = Q / 4πϵ₀R = V_S

When x = 2R, V = Q / 4πϵ₀(2R) = ½ V_S           (2R, 0.5 V_S)  

When x = 3R, V = Q / 4πϵ₀(3R) = 1/3 V_S        (2R, 0.33 V_S)  

}

(b)

sketch: from x = 0 to x = R, field strength is zero

smooth curve through (R, E) and (2R, 0.25E)

smooth curve continues to (3R, 0.11E)

{Electric field strength: E = Q / 4πϵ₀x²           where x is the distance from the centre

For a conductor, the field strength is zero inside the sphere. So, from x = 0 to x = R (that is, inside the sphere), field strength = zero. This is a horizontal line in the graph.

At the surface of the sphere, the field strength is maximum. This is the point (R, 1.0 E_S).

           

From the equation for field strength above, the field strength E is inversely proportional to x². The greater the value of x, the smaller the potential.

E = Q / 4πϵ₀x²

When x = R, E = Q / 4πϵ₀R² = E_S                 (R, E_S)

When x = 2R, E = Q / 4πϵ₀(2R)² = ¼ E_S       (R, 0.25E_S)

When x = 3R, E = Q / 4πϵ₀(3R)² = 1/9 E_S     (R, 0.11E_S)

}