The work done in moving a $500\mu C$ charge between two points on equipotential surface is:
(A) Zero
(B) Finite positive
(C) Finite negative
(D) Infinite

Hint: To solve this question, we have to consider the meaning of the term equipotential. Then, considering any two points on the given equipotential surface, we can find out the value of the work done by using its formula.
Formula used: The formula used to solve this question is given by
$W = Q\Delta V$, here $W$ is the work done in moving a charge of $Q$ between two points which have a potential difference of $\Delta V$.

Complete step by step solution:
We know that an equipotential surface is a collection of all points in the space, which have the same potential. It is different for different kinds of charge distributions.
Let us name the two points on the given equipotential surface, between which the given charge is moved, as A and B.
Since the given surface is equipotential, so according to the above definition, the potentials of the points A and B will be the same, that is,
${V_A} = {V_B}$
\[ \Rightarrow {V_B} - {V_A} = 0\] ………………..(1)
Now, we know that the work done in moving a point is given by
$W = Q\Delta V$
$ \Rightarrow W = Q\left( {{V_B} - {V_A}} \right)$ …………………...(2)
According to the question, the value of the given charge is
$Q = 500\mu C$
We know that $1\mu C = {10^{ - 6}}C$. So the value of the given charge becomes
$Q = 500 \times {10^{ - 6}}C$
On simplifying we get
\[Q = 5 \times {10^{ - 4}}C\] ……………………..(3)
Substituting (3) in (2), we get
$ \Rightarrow W = 5 \times {10^{ - 4}}\left( {{V_B} - {V_A}} \right)$
Substituting (1), we finally get
$W = 0$
So the work done in moving the given charge between two equipotential surfaces is equal to zero.

Hence, the correct answer is option A.

Note: When we talk of moving a charged particle between two points in electrostatics, it is assumed that there should be no change in the kinetic energy of the particle. This rule is followed unless stated otherwise. This assumption is made in order to keep our analysis limited to the change in potential energy only.