Question 30
Two small solid metal spheres A and B have equal radii and are in a vacuum. Their centres are 15 cm apart.
Sphere A has charge +3.0 pC and sphere B has charge +12 pC. The arrangement is illustrated in Fig. 5.1.
Fig. 5.1
Point P lies on the line joining the centres of the spheres and is a distance of 5.0 cm from the centre of sphere A.
(a) Suggest why the electric field strength in both spheres is zero. [2]
(b) Show that the electric field strength is zero at point P. Explain your working. [3]
(c) Calculate the electric potential at point P. [2]
(d) A silver-107 nucleus (10747 Ag) has speed v when it is a long distance from point P.
Use your answer in (c) to calculate the minimum value of speed v such that the nucleus can reach point P. [3]
[Total: 10]
Reference: Past Exam Paper – November 2016 Paper 41 & 43 Q5
Solution:
(a)
In an electric field, the charges (in a conductor) would move.
Since there is no movement of charge inside the sphere, the field strength is zero.
(b)
For the electric field strength to be zero at P, the fields due to the charged spheres must be (equal and) opposite in direction, so that E = 0.
{Showing that the field strength due to each sphere are equal in magnitude:
Electric field strength, E = Q / 4πϵ0r2
where r is the distance of P from the centre of the sphere
For sphere A, r = 5 cm = 0.05 m
For sphere B, r = 15 – 5 = 10 cm = 0.10 m
Note: 1 pC = 1×10-12 C}
at P, EA = (3.0×10-12) / [4 πϵ0 × (5.0×10-2)2] (= 10.79 N C-1)
at P, EB = (12×10-12) / [4 πϵ0 × (10×10-2)2] (= 10.79 N C-1)
(c)
{Electric potential V = Q / 4πϵ0r
At point P, the electric potential is due to both spheres A and B. Since electric potential is a scalar, we add both potentials directly (no directions are involved).
Electric potential at P = VA + VB
Electric potential at P = (QA / 4πϵ0rA) + (QB / 4πϵ0rB)
Electric potential at P = 1/4πϵ0 × [(QA / rA) + (QB / rB)]
1/4πϵ0 = 8.99×109 mF-1 as given in the list of data in the question paper
Electric potential at P = 8.99×109 × [(QA / rA) + (QB / rB)]}
potential = 8.99×109 × {(3.0×10-12) / (5.0×10-2) + (12×10-12) / (10×10-2)}
potential = 1.62 V
(d)
{Kinetic energy of nucleus = Electrical energy
Electrical energy = qV where q is the charge of the nucleus}
½mv2 = qV
{The silver nucleus contains 107 nucleons. So, its mass is 107u (= 107 × 1.66×10-27 kg).}
EK = ½ ×107 × 1.66×10-27 × v2
{The silver nucleus contains 47 protons. So, its charge is 47e (=47 × 1.60×10-19 C).}
qV = 47 × 1.60×10-19 × 1.62
v2 = 1.37 × 108
v = 1.2 × 104 m s-1
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