
The work done on a charge when it moves along an equipotential surface is zero. This is because the potential difference between any two points on the same equipotential surface is zero. To understand this concept, let's look at the work done when moving a charge in the presence of a potential difference.
The work done \( W \) when moving a charge \( Q \) between two points with potential difference \( V_1 \) and \( V_2 \) is given by:
\[ W = Q(V_2 - V_1) \]
For any path along an equipotential surface, \( V_1 = V_2 \), meaning there is no potential difference, and therefore the work done \( W \) is zero.
From the given figure, the maximum potential difference occurs when the charge moves from the 25V surface to the 10V surface. The path corresponding to this maximum potential difference is along path D. Since the potential difference is the largest between these two surfaces, the maximum work done will be along this path.
Thus, the correct answer is path D, as this gives the maximum potential difference and the maximum work done.
Match List-I with List-II.
Choose the correct answer from the options given below :}
There are three co-centric conducting spherical shells $A$, $B$ and $C$ of radii $a$, $b$ and $c$ respectively $(c>b>a)$ and they are charged with charges $q_1$, $q_2$ and $q_3$ respectively. The potentials of the spheres $A$, $B$ and $C$ respectively are:
Two resistors $2\,\Omega$ and $3\,\Omega$ are connected in the gaps of a bridge as shown in the figure. The null point is obtained with the contact of jockey at some point on wire $XY$. When an unknown resistor is connected in parallel with $3\,\Omega$ resistor, the null point is shifted by $22.5\,\text{cm}$ towards $Y$. The resistance of unknown resistor is ___ $\Omega$. 

