Question

Statement 1: Charge density on a solid conductor is always uniform no matter what the shape.
Statement 2: Charges are always on the surface of a conductor and they repel each other so they always arrange to give uniform charge density.

• A. Statement 1 is true. Statement 2 is false
• B. Statement 1 is false. Statement 2 is true
• C. Both statements are true
• D. Both statements are false

Answer 1

The correct option is D Both statements are false

The only thing that remains constant throughout a conductor surface is potential. Let’s see what this means for charge density.
We know that potential of a uniformly charged sphere of radius $r_1$ having charge $Q_1$ is $V_1=\frac{Q_1}{4\pi {\epsilon }_0{r}_1}$
The same for another sphere of smaller radius $r_2$ and charge $Q_2$ is $V_2=\frac{Q_2}{4\pi {\epsilon }_0{r}_2}$
If these two spheres are connected either by a wire (or are part of the same solid conductor),
Then, we can equate the potentials to get,
$\frac{Q_1}{4\pi {\epsilon }_0{r}_1}=\frac{Q_2}{4\pi {\epsilon }_0{r}_2}$
$\frac{Q_1\times r_1}{4\pi {\epsilon }_0{r}_1\times r_1}=\frac{Q_2\times r_2}{4\pi {\epsilon }_0{r}_2\times r_2}$
Since charge by surface area gives charge density $\sigma$,
${\sigma }_1{r}_1={\sigma }_2{r}_2$
$\frac{{\sigma }_1}{{\sigma }_2}=\frac{r_2}{r_1}$
The higher the radius of curvature of a surface, the lower the charge density. Thus the sharper parts of a conductor will have higher charge density. And the broader parts will have less charge density. This is what leads to Corona discharge.
Statement 1 is obviously false.
With regard to statement 2, all that can be said is that, just because like charges repel, doesn’t mean they arrange to give uniform charge density. They will arrange till they are in equilibrium and that gives varying densities on irregularly shaped conductors