Question

A point charge causes an electric flux of $-1.0\times10^{3}\,\text{N m}^{2}/\text{C}$ to pass through a spherical Gaussian surface of $10.0\,\text{cm}$ radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?

Hint:

Flux depends only on enclosed charge, so doubling the radius leaves it unchanged. Then q = epsilon_0 times flux.

Answer 1

Step 1: Recall Gauss's law: the flux through a closed surface depends only on the enclosed charge:
\[\Phi = \frac{q}{\epsilon_0}\]
Step 2 (part a): If the radius of the spherical Gaussian surface is doubled, the same point charge is still the only charge enclosed. Since the flux does not depend on the radius, it stays the same:
\[\Phi' = \Phi = -1.0\times10^{3}\,\text{N m}^2\,\text{C}^{-1}\]
Step 3 (part b): Find the charge from \(\Phi = q/\epsilon_0\), so \(q = \epsilon_0\,\Phi\). Substitute the values:
\[q = (8.85\times10^{-12})\times(-1.0\times10^{3})\]
Step 4: Do the arithmetic:
\[q = -8.85\times10^{-9}\,\text{C} = -8.85\,\text{nC}\]
The negative sign shows the charge is negative (consistent with inward flux).
\[\boxed{\Phi' = -1.0\times10^{3}\,\text{N m}^2\,\text{C}^{-1},\quad q = -8.85\,\text{nC}}\]

Answer 2

Field-and-area reasoning (alternate framing).
Step 1: The field of a point charge at radius \(r\) is \(E = \dfrac{q}{4\pi\epsilon_0 r^2}\), and the area of the sphere is \(A = 4\pi r^2\). The flux is the product:
\[\Phi = E\,A = \frac{q}{4\pi\epsilon_0 r^2}\times 4\pi r^2 = \frac{q}{\epsilon_0}\]
Step 2 (part a): When \(r \to 2r\), the field falls by a factor of 4 (\(1/r^2\)) but the area grows by a factor of 4 (\(r^2\)). The two changes cancel, so the flux is unchanged:
\[\Phi' = -1.0\times10^{3}\,\text{N m}^2\,\text{C}^{-1}\]
Step 3 (part b): Solve \(\Phi = q/\epsilon_0\) for the charge:
\[q = \epsilon_0\,\Phi = (8.85\times10^{-12})(-1.0\times10^{3}) = -8.85\times10^{-9}\,\text{C}\]
\[\boxed{\Phi' = -1.0\times10^{3}\,\text{N m}^2\,\text{C}^{-1},\quad q = -8.85\,\text{nC}}\]