Question

A square metal sheet of 4 m \( \times \) 4 m is placed on the x-y plane as shown in the figure below. If the surface charge density (in \( \mu \)C/m\(^2\)) on the sheet is \( \rho_s(x, y) = 4|y| \), then the total charge (in \( \mu \)C, rounded off to the nearest integer) on the sheet is _________.

• A. 16
• B. 85
• C. 64
• D. 256

Hint:

To calculate the total charge from surface charge density, set up a double integral over the area, carefully considering the sign of the charge density.

Answer 1

The Correct Option is C

The total charge on the sheet is given by the integral of the surface charge density \( \rho_s(x, y) \) over the area of the sheet: \[ Q = \int \int_{A} \rho_s(x, y) \, dA. \] The area of the square sheet is 4 m \( \times \) 4 m, so the limits for \( x \) and \( y \) are from \( -2 \, {m} \) to \( 2 \, {m} \). The surface charge density is \( \rho_s(x, y) = 4|y| \), so: \[ Q = \int_{-2}^{2} \int_{-2}^{2} 4|y| \, dx \, dy. \] First, integrate with respect to \( x \): \[ Q = \int_{-2}^{2} 4|y| \cdot (2 - (-2)) \, dy = \int_{-2}^{2} 16|y| \, dy. \] Now, integrate with respect to \( y \): \[ Q = 16 \cdot \int_{-2}^{2} |y| \, dy = 16 \cdot 2 \cdot \int_{0}^{2} y \, dy = 16 \cdot 2 \cdot \left[ \frac{y^2}{2} \right]_0^2. \] This gives: \[ Q = 16 \cdot 2 \cdot \frac{4}{2} = 64 \, \mu{C}. \] Thus, the correct answer is (C).