Question: A positive point charge is placed near a very large conducting plane. A professor of physics asserted that the field caused by this configuration is the same as would be obtained by removing the plane and placing a negative point charge of equal magnitude in the mirror image position behind the initial position of the plane. Is this correct? Why or why not?

The term electric potential is defined as the amount of work done by the unit charge in moving from one point to another against electric field.

The electric field obtained by the relation:

$\overrightarrow{E}=-\overrightarrow{\nabla }V$

Consider the electric potential between the two point is as follows:

Taking integration on side and taking limit from

$$\begin{gathered} {\int }_A^{B}dV=-\int Edr \\ {V}_B-V_A=-\int 0dr+C \\ {V}_B-V_A=C \\ {V}_B-V_A=C \end{gathered}$$

The atom on surface of plate consists positive nucleus and negative electrons. Near the surface of plate positive charge is there due to Coulomb’s law and the outer surface electron attracts the positive charge. While the plate surface is equipotential. The center of the plate can be replaced by the negative charge plate. In both case electric field line will be same. And in both cases electric potential is also same.

Hence, it is correct. In both cases electric potential is same.