Q#24 (Past Exam Paper – November 2014 Paper 41 & 42 Q5)
(a) Define electric potential at a point. [2]
(b) An isolated solid metal sphere is positively charged.
The variation of the potential V with distance x from the centre of the sphere is shown in
Fig. 5.1.
Fig. 5.1
Use Fig. 5.1 to suggest
(i) why the radius of the sphere cannot be greater than 1.0 cm, [1]
(ii) that the charge on the sphere behaves as if it were a point charge. [3]
(c) Assuming that the charge on the sphere does behave as a point charge, use data from Fig. 5.1 to determine the charge on the sphere. [2]
Solution:
(a) The electric potential at a point is defined as the work done in moving unit positive charge from infinity (to the point).
(b)
(i) The electric potential is constant inside the sphere
{Electric potential decreases from the surface of the sphere as we go further away. BUT inside the sphere the electric potential should be constant (as the electric field strength is zero inside the sphere).
Since the graph starts at x = 1.0 cm it can be deduced that any distance smaller than x = 1.0 cm is inside the sphere.
If the radius was greater than 1.0 cm, the electric potential should have been constant for some values of x greater than 1.0 cm. However, we can observe that the potential decreases as from x = 1.0 cm. So the radius cannot be greater than 1.0 cm.}
(ii)
{V = Q / 4πε0x
Since 4πε0 is constant,
V ∝ Q / x
Vx ∝ Q
The product of Vx is proportional to the charge Q.
For a point charge, the value of Q should be constant – that is, the charge cannot be changing.}
For a point charge, the product Vx is constant.
{From the graph, a point represents (x, V). So for any point on the graph, the product of Vx should be given the same (constant) value.
We need to use points from the graph to show that the products for these points are constant.}
The co-ordinates should be clear and we need to determine two values of Vx at least 4 cm apart.
{Consider the points (6, 30) and (2, 90) from the graph. (the points cannot be too close)
The product of both of them give a value of 180 (6×30 = 180 and 2×90 = 180). So, the product obtained is constant at different points.}
Since the product is constant, the charge on the sphere behaves as if it were a point charge.
(c)
{As found above, the product is 180 Vcm
Vx = 180 Vcm = 180 × (1.0×10–2) Vm
V = Q / 4πε0x
Q = V×4πε0x = 4πε0[Vx] }
Charge Q = 4πε0[Vx] = 4π × (8.85×10–12) × [180 × (1.0×10–2)] = 2.0×10–10 C
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