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Question

A conducting sphere has radius 48 cm. The electric potential on the surface of the sphere is 3.4 × 10⁵ V.

The sphere is connected by a long conducting wire to a second conducting sphere of radius 24 cm. The second sphere is initially uncharged.

Show that the charge on the surface of the sphere is +18 μC.

[1]

Describe, in terms of electron flow, how the smaller sphere becomes charged.

[1]

b.i.

Predict the charge on each sphere.

[3]

b.ii.

$Q = « \frac{V R}{k} = » \frac{3.4 \times 10^{5} \times 0.48}{8.99 \times 10^{9}}$

$Q = 18.2$ «μC» ✓

electrons leave the small sphere «making it positively charged» ✓

b.i.

$k \frac{q_{1}}{48} = k \frac{q_{2}}{24} \Rightarrow q_{1} = 2 q_{2}$ ✓

$q_{1} + q_{2} = 18$ ✓

so $q_{1} = 12$ «μC», $q_{2} = 6.0$ «μC» ✓

Award [3] marks for a bald correct answer.

b.ii.

[N/A]

b.i.

[N/A]

b.ii.