A thin square plate is placed in the x-y plan | vedclass

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(D) The total charge $Q$ on the plate is given by the surface integral of the charge density $\sigma$ over the area of the plate:$Q = \int_{-a}^{a} \i

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Zero

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(D) The total charge $Q$ on the plate is given by the surface integral of the charge density $\sigma$ over the area of the plate:$Q = \int_{-a}^{a} \int_{-a}^{a} \sigma_0 xy \, dx \, dy$$Q = \sigma_0 \left( \int_{-a}^{a} x \, dx ight) \left( \int_{-a}^{a} y \, dy ight)$Since the integral of an odd function over a symmetric interval $[-a, a]$ is zero:$\int_{-a}^{a} x \, dx = 0$ and $\int_{-a}^{a} y \, dy = 0$Therefore,$Q = \sigma_0 	imes 0 	imes 0 = 0$.Alternatively,by symmetry,the charge in the first quadrant is positive,the second is negative,the third is positive,and the fourth is negative. Due to the symmetry of the function $\sigma = \sigma_0 xy$ about the axes,the total charge sums to zero.

$A$ thin square plate is placed in the $x-y$ plane as shown in the figure,such that its center coincides with the origin. Its charge density at point $(x, y)$ is $\sigma = \sigma_0 xy$ (where $\sigma_0$ is a constant). Find the total charge on the plate.

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