Question

If four charges \( q_1 = +1 \times 10^{-8} C \), \( q_2 = -2 \times 10^{-8} C \), \( q_3 = +3 \times 10^{-8} C \), and \( q_4 = +2 \times 10^{-8} C \) are kept at the four corners of a square of side 1 m, then the electric potential at the centre of the square is:

• A. \( 300 \) V
• B. \( 200 \) V
• C. \( 510 \) V
• D. \( 410 \) V

Hint:

The electric potential due to multiple charges is algebraic sum because potential is a scalar quantity. Use \( V = \frac{kq}{r} \) for each charge and sum them.

Answer 1

The Correct Option is C

Step 1: Using the Formula for Electric Potential The electric potential at the center due to a charge \( q \) placed at a distance \( r \) is given by: \[ V = \frac{kq}{r}. \] Since all four charges are at the same distance \( r = \frac{1}{\sqrt{2}} \) m from the center, the total potential is: \[ V_{\text{total}} = k \left( \frac{q_1 + q_2 + q_3 + q_4}{r} \right). \]
Step 2: Substituting Values \[ V_{\text{total}} = 9 \times 10^9 \times \left( \frac{(1 - 2 + 3 + 2) \times 10^{-8}}{1/\sqrt{2}} \right). \] \[ V_{\text{total}} = 9 \times 10^9 \times \left( \frac{4 \times 10^{-8}}{1/\sqrt{2}} \right). \] \[ V_{\text{total}} = 510 V. \] Thus, the correct answer is: \[ \boxed{510 \text{ V}}. \]

Answer 2

The electric potential \( V \) at a point due to a point charge is given by the formula:

\( V = \frac{k \cdot q}{r} \)

where \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) (Coulomb's constant), \( q \) is the charge, and \( r \) is the distance from the charge to the point of interest.

Given four charges \( q_1 = +1 \times 10^{-8} \, \text{C} \), \( q_2 = -2 \times 10^{-8} \, \text{C} \), \( q_3 = +3 \times 10^{-8} \, \text{C} \), \( q_4 = +2 \times 10^{-8} \, \text{C} \) at the corners of a square with side length 1 m, the distance from each charge to the center of the square is:

\( r = \frac{\text{diagonal}}{2} = \frac{\sqrt{2}}{2} \, \text{m} \)

The electric potential at the center \( V_c \) is the sum of potentials due to each charge:

\( V_c = V_1 + V_2 + V_3 + V_4 \)

Compute the potential from each charge:

• \( V_1 = \frac{k \cdot q_1}{r} = \frac{9 \times 10^9 \times 1 \times 10^{-8}}{\sqrt{2}/2} = \frac{18 \times 10}{\sqrt{2}} \)
• \( V_2 = \frac{k \cdot q_2}{r} = \frac{9 \times 10^9 \times (-2) \times 10^{-8}}{\sqrt{2}/2} = \frac{-36 \times 10}{\sqrt{2}} \)
• \( V_3 = \frac{k \cdot q_3}{r} = \frac{9 \times 10^9 \times 3 \times 10^{-8}}{\sqrt{2}/2} = \frac{54 \times 10}{\sqrt{2}} \)
• \( V_4 = \frac{k \cdot q_4}{r} = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{\sqrt{2}/2} = \frac{36 \times 10}{\sqrt{2}} \)

Summing these potentials, we have:

\( V_c = \frac{18 \times 10 + (-36 \times 10) + 54 \times 10 + 36 \times 10}{\sqrt{2}} = \frac{72 \times 10}{\sqrt{2}} = 510 \, \text{V} \)

Therefore, the electric potential at the center of the square is \( 510 \) V.