Return to Answer

As you say, you should use Gauss's law. You pick a Gaussian surface that is a sphere concentric with the insulated sphere, with radius less than or equal to $R$. Then you get:

$$ 4\pi r^2 E = \frac{Q_{enc}(r)}{\epsilon_0} $$

with $Q_{enc}(r)$ the charge enclosed by the surface of radius $r$.

If you pick $r = R$ you get $Q_{enc}(R) = 4\pi R^2E\epsilon_0$.

I'm not entirely comfortable that this is consistent with a physical distribution of charge. Because $E$ is constant everywhere, it must be that $Q_{enc}(r) \propto r^2$. I can't immediately find a charge distribution which achieves this without requiring infinite charge density at the origin. I'm not saying it's wrong, but something to be cautious about.

As you say, you should use Gauss's law. You pick a Gaussian surface that is a sphere concentric with the insulated sphere, with radius less than or equal to $R$. Then you get:

$$ 4\pi r^2 E = \frac{Q_{enc}(r)}{\epsilon_0} $$

with $Q_{enc}(r)$ the charge enclosed by the surface of radius $r$.

If you pick $r = R$ you get $Q_{enc}(R) = 4\pi R^2E\epsilon_0$.

I'm not entirely comfortable that this is consistent with a physical distribution of charge. Because $E$ is constant everywhere, it must be that $Q_{enc}(r) \propto r^2$. But then charge density is infinite at the centre.

As you say, you should use Gauss's law. You pick a Gaussian surface that is a sphere concentric with the insulated sphere, with radius less than or equal to $R$. Then you get:

$$ 4\pi r^2 E = \frac{Q_{enc}}{\epsilon_0} $$

with $Q_{enc}$ the charge enclosed by the surface.

If you pick $r = R$ you get $Q_{enc} = 4\pi R^2E\epsilon_0$.

As you say, you should use Gauss's law. You pick a Gaussian surface that is a sphere concentric with the insulated sphere, with radius less than or equal to $R$. Then you get:

$$ 4\pi r^2 E = \frac{Q_{enc}(r)}{\epsilon_0} $$

with $Q_{enc}(r)$ the charge enclosed by the surface of radius $r$.

If you pick $r = R$ you get $Q_{enc}(R) = 4\pi R^2E\epsilon_0$.

I'm not entirely comfortable that this is consistent with a physical distribution of charge. Because $E$ is constant everywhere, it must be that $Q_{enc}(r) \propto r^2$. I can't immediately find a charge distribution which achieves this without requiring infinite charge density at the origin. I'm not saying it's wrong, but something to be cautious about.

As you say, you should use Gauss's law. You pick a Gaussian surface that is a sphere concentric with the insulated sphere, with radius less than or equal to $R$. Then you get:

$$ 4\pi r^2 E = \frac{Q_{enc}}{\epsilon_0} $$

with $Q_{enc}$ the charge enclosed by the surface.

If you pick $r = R$ you get $Q_{enc} = 4\pi R^2E\epsilon_0$.

However, I think there is possibly a problem with this solution. Rearranging the general expression, you get:

$$ E = \frac{Q_{enc}}{4\pi r^2 \epsilon_0} $$

and as $r$ goes to $0$, $Q_{enc}$ needs to get larger and larger to ensure that $E$ is kept constant. The implication is that there is infinite charge at the centre of the sphere, which is not physical.

Things would be easier if $E$ was only constant on the surface of the sphere, and not all the way through the sphere.

As you say, you should use Gauss's law. You pick a Gaussian surface that is a sphere concentric with the insulated sphere, with radius less than or equal to $R$. Then you get:

$$ 4\pi r^2 E = \frac{Q_{enc}}{\epsilon_0} $$

with $Q_{enc}$ the charge enclosed by the surface.

If you pick $r = R$ you get $Q_{enc} = 4\pi R^2E\epsilon_0$.