Question

A plane square sheet of charge of side 0.5 m has uniform surface charge density. An electron at 1 cm from the centre of the sheet experiences a force of \( 1.6 \times 10^{-12} \) N directed away from the sheet. The total charge on the plane square sheet is (\( \epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{m}^{-2}\text{N}^{-1} \))

• A. \( 16.25 \text{ } \mu\text{C} \)
• B. \( -22.15 \text{ } \mu\text{C} \)
• C. \( -44.27 \text{ } \mu\text{C} \)
• D. \( 144.27 \text{ } \mu\text{C} \)
• E. \( 8.854 \text{ } \mu\text{C} \)

Hint:

For large sheets, the electric field is considered uniform (\(E = \sigma/2\epsilon_0\)) at distances much smaller than the dimensions of the sheet (1 cm vs 50 cm). This makes the field independent of the distance!

Answer 1

The Correct Option is C

Concept: This problem involves the electric field of an infinite plane sheet of charge and the force on a point charge.
• Electric Field (\(E\)): For an infinite sheet, \( E = \frac{\sigma}{2\epsilon_0} \), where \(\sigma\) is surface charge density.
• Electrostatic Force: \( F = qE \). For an electron, \( q = -1.6 \times 10^{-19} \text{ C} \).
• Total Charge (\(Q\)): \( Q = \sigma \times \text{Area} \).

Step 1: Determine the sign of the charge.
The electron (negative charge) experiences a force away from the sheet. Since opposite charges attract and like charges repel, the sheet must be negatively charged to push the electron away. This eliminates options A, D, and E.

Step 2: Calculate the Electric Field \(E\).
From \( F = qE \): \[ E = \frac{F}{e} = \frac{1.6 \times 10^{-12} \text{ N}}{1.6 \times 10^{-19} \text{ C}} = 10^7 \text{ N/C} \]

Step 3: Solve for total charge \(Q\).
Using \( E = \frac{\sigma}{2\epsilon_0} \) and \(\sigma = \frac{Q}{A}\): \[ E = \frac{Q}{2A\epsilon_0} \implies Q = 2E A \epsilon_0 \] Area of the square sheet \( A = (0.5 \text{ m})^2 = 0.25 \text{ m}^2 \). \[ Q = 2 \times 10^7 \times 0.25 \times 8.854 \times 10^{-12} \] \[ Q = 0.5 \times 8.854 \times 10^{-5} = 4.427 \times 10^{-5} \text{ C} \] Converting to microCoulombs (\(\mu\text{C}\)) and applying the negative sign: \[ Q = -44.27 \text{ } \mu\text{C} \]