(a) Calculation:

Solve for the new charge density on the outside of the sphere:

As equation (22.10) mentions, the electric field at the surface of a conductor is given by

$E=\frac{\sigma}{{\u03f5}_{0}}$

We use the same technique as an example (22.5), the electric field at the surface of a charged conducting sphere is given by

$E=\frac{1}{4\pi {\u03f5}_{0}}\frac{Q}{{r}^{2}}$

Substituting from the previous calculation and solve for the charge density on the outside of the sphere, then we get

$\sigma}_{out}=E{\u03f5}_{0$

$=\frac{1}{4\pi {\u03f5}_{0}}\frac{Q}{{r}^{2}}{\u03f5}_{0}$

$=\frac{Q}{4\pi {r}^{2}}$

$=\frac{-0.5\times {10}^{-6}C}{4\pi {\left(0.25m\right)}^{2}}$

$=-6.366\times {10}^{-7}\frac{C}{{m}^{2}}$ Calculation:

In order to evaluate the new charge density on the outside of the sphere, we use the following relation:

$\sigma}_{\ne w}={\sigma}_{\in}-{\sigma}_{out$

$=-6.37\times {10}^{-6}\frac{C}{{m}^{2}}-6.366\times {10}^{-7}\frac{C}{{m}^{2}}$

$=5.733\times {10}^{-6}\frac{C}{{m}^{2}}$

So, the new charge density on the outside of the sphere is $5.733\times {10}^{-6}\frac{C}{{m}^{2}}$