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(D) The square plate has side length $a$ and is centered at the origin $(0,0)$ in the $xy$-plane. The limits for $x$ and $y$ are from $-\frac{a}{2}$ t

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Zero

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(D) The square plate has side length $a$ and is centered at the origin $(0,0)$ in the $xy$-plane. The limits for $x$ and $y$ are from $-\frac{a}{2}$ to $+\frac{a}{2}$.The total charge $Q$ is given by the surface integral of the charge density $\sigma$ over the area of the plate:$Q = \iint \sigma \, dA = \int_{-\frac{a}{2}}^{\frac{a}{2}} \int_{-\frac{a}{2}}^{\frac{a}{2}} (xy) \, dx \, dy$This can be separated into two independent integrals:$Q = \left( \int_{-\frac{a}{2}}^{\frac{a}{2}} x \, dx ight) \left( \int_{-\frac{a}{2}}^{\frac{a}{2}} y \, dy ight)$Since the integrand $x$ is an odd function and the limits are symmetric about the origin,$\int_{-\frac{a}{2}}^{\frac{a}{2}} x \, dx = 0$.Similarly,$\int_{-\frac{a}{2}}^{\frac{a}{2}} y \, dy = 0$.Therefore,$Q = 0 	imes 0 = 0$.The total charge on the plate is zero.

$A$ square plate of side $a$ is placed in the $xy$-plane with its center at the origin. If the surface charge density of the square plate is given by $\sigma = xy$,then the total charge on the plate will be:

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