The work done in moving a charge \( 2 \times 10^{-9} \mathrm{C} \) from a point of potential -3 kV to another poin \( P \) is \( 5 \times 10^{-5} \mathrm{~J} \). Find the potential at point \( P \)

To find the potential at point \( P \), you can use the relationship between work done, charge, and potential difference. The formula is given by: \[ W = q \cdot (V_P - V_i) \] Where: - \( W \) is the work done (in joules), - \( q \) is the charge (in coulombs), - \( V_P \) is the potential at point \( P \), - \( V_i \) is the initial potential (-3000 V here). Given: - Work done, \( W = 5 \times 10^{-5} \mathrm{~J} \) - Charge, \( q = 2 \times 10^{-9} \mathrm{C} \) - Initial potential, \( V_i = -3000 \mathrm{~V} \) We can rearrange the equation to solve for \( V_P \): \[ V_P = V_i + \frac{W}{q} \] Now, substituting in the known values: \[ V_P = -3000 + \frac{5 \times 10^{-5}}{2 \times 10^{-9}} \] Calculating the second term: \[ \frac{5 \times 10^{-5}}{2 \times 10^{-9}} = 25000 \mathrm{~V} \] Now substitute back into the equation for \( V_P \): \[ V_P = -3000 + 25000 = 22000 \mathrm{~V} \] Thus, the potential at point \( P \) is \( 22000 \mathrm{~V} \) or \( 22 \mathrm{kV} \).