Question

A conducting sphere of radius $10\,\text{cm}$ has an unknown charge. If the electric field $20\,\text{cm}$ from the centre of the sphere is $1.5\times10^{3}\,\text{N/C}$ and points radially inward, what is the net charge on the sphere?

Hint:

Outside the sphere treat the charge as a point charge: q = E r^2 / k. Inward field means the charge is negative.

Answer 1

Step 1: Outside a charged conducting sphere, the field is the same as if all its charge were concentrated at the centre:
\[E = \frac{1}{4\pi\epsilon_0}\,\frac{q}{r^2} = \frac{k\,q}{r^2}\]
Step 2: Rearrange to find the charge, using \(k = 9\times10^{9}\,\text{N m}^2\,\text{C}^{-2}\):
\[q = \frac{E\,r^2}{k}\]
Step 3: Substitute \(E = 1.5\times10^{3}\,\text{N C}^{-1}\) and \(r = 20\,\text{cm} = 0.20\,\text{m}\):
\[q = \frac{(1.5\times10^{3})\times(0.20)^2}{9\times10^{9}}\]
Step 4: Do the arithmetic:
\[q = \frac{(1.5\times10^{3})\times(0.04)}{9\times10^{9}} = \frac{60}{9\times10^{9}} = 6.67\times10^{-9}\,\text{C}\]
Step 5: Fix the sign. The field points radially inward (toward the centre), which happens only for a negative charge. So:
\[q = -6.67\times10^{-9}\,\text{C} = -6.67\,\text{nC}\]
The radius of the sphere (10 cm) is not needed because the field point lies outside it.
\[\boxed{q = -6.67\,\text{nC}}\]

Answer 2

Gauss's law route.
Step 1: Take a spherical Gaussian surface of radius \(r = 0.20\,\text{m}\) concentric with the conductor. By Gauss's law:
\[E\,(4\pi r^2) = \frac{q}{\epsilon_0}\]
Step 2: Solve for the magnitude of the charge:
\[q = E\,(4\pi\epsilon_0)\,r^2\]
Step 3: Insert \(4\pi\epsilon_0 = 1/k = 1.11\times10^{-10}\), \(E = 1.5\times10^{3}\), \(r^2 = 0.04\):
\[q = (1.5\times10^{3})\times(1.11\times10^{-10})\times(0.04)\]
Step 4: Evaluate:
\[q = 6.67\times10^{-9}\,\text{C}\]
Step 5: An inward radial field means flux enters the surface, so the enclosed charge is negative:
\[q = -6.67\times10^{-9}\,\text{C}\]
\[\boxed{q = -6.67\,\text{nC}}\]