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.
Bittu and Chintu were partners in a firm sharing profit and losses in the ratio of 4:3. Their Balance Sheet as at 31st March, 2024 was as
On $1^{\text {st }}$ April, 2024, Diya was admitted in the firm for $\frac{1}{7}$ share in the profits on the following terms:
Prepare Revaluation Account and Partners' Capital Accounts.
(a) Calculate the standard Gibbs energy (\(\Delta G^\circ\)) of the following reaction at 25°C:
\(\text{Au(s) + Ca\(^{2+}\)(1M) $\rightarrow$ Au\(^{3+}\)(1M) + Ca(s)} \)
\(\text{E\(^\circ_{\text{Au}^{3+}/\text{Au}} = +1.5 V, E\)\(^\circ_{\text{Ca}^{2+}/\text{Ca}} = -2.87 V\)}\)
\(\text{1 F} = 96500 C mol^{-1}\)
Define the following:
(i) Cell potential
(ii) Fuel Cell