Question

1) For equipotential surfaces the following statement is correct or not:

A) Equipotential surfaces are closer in regions of large electric fields compared to regions of lower electric fields

2) For equipotential surfaces the following statement is correct or not:

B) Equipotential surfaces will be more crowded near sharp edges of a conductor

3) For equipotential surfaces the following statement is correct or not:

C) Equipotential surfaces will be more crowded near regions of large charge densities

4) For equipotential surfaces the following statement is correct or not:

D) Equipotential surfaces will always be equally spaced

Answer 1

1)

The electric field intensity is given by

$E=\frac{-dV}{dr}$

For any two points of equipotential surface, potential difference is zero.

So, $E\alpha \frac{1}{r}$

Therefore, equipotential surfaces are closer in regions of large electric fields.

Final Answer: Correct

2)

As we know that the surface charge density is given by

$\because \sigma =\frac{q}{A}$

The electric field is larger near the sharp edge, due to larger charge density as $A$ is very small.

So equipotential surfaces are closer or crowded.

Final Answer: Correct

3)

The electric field intensity is given by

$E=\frac{kq}{r^2}$

$\therefore r\uparrow \Rightarrow E\downarrow$

Or potential or field decreases as size of body increases or vice-versa (case of earth).

And from the formula $\sigma =\frac{q}{A}$

The equipotential surfaces will be more crowded if the charge density $\left(\sigma \right)$ increases.

Final Answer: Correct

4)

The distance between equipotential surface depends on distance $r$.

Electric field intensity is given by

$E=-\frac{dV}{dr}$

and potential $V=\frac{kq}{r}$

It is different at different place. Equipotential surface are not equispaced all over.

Final Answer: Incorrect