Four point charges -q, +q, +q and -q are plac | vedclass

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(D) The electric field at point $P(D, 0)$ is the vector sum of the fields due to the four charges. Due to symmetry,the $y$-components of the electric

Vedclass · Accepted Answer

$E \propto \frac{1}{D^4}$

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(D) The electric field at point $P(D, 0)$ is the vector sum of the fields due to the four charges. Due to symmetry,the $y$-components of the electric fields cancel out. The net electric field is along the $x$-axis.The electric field due to the pair of charges $+q$ at $y = \pm d$ is $E_1 = \frac{2kq}{(d^2+D^2)} \cos 	heta_1 = \frac{2kqD}{(d^2+D^2)^{3/2}}$.The electric field due to the pair of charges $-q$ at $y = \pm 2d$ is $E_2 = \frac{2kq}{((2d)^2+D^2)} \cos 	heta_2 = \frac{2kqD}{((2d)^2+D^2)^{3/2}}$.The net field $E = E_1 - E_2 = 2kqD \left[ (d^2+D^2)^{-3/2} - (4d^2+D^2)^{-3/2} ight]$.Factoring out $D^2$ from the brackets: $E = \frac{2kqD}{D^3} \left[ (1 + \frac{d^2}{D^2})^{-3/2} - (1 + \frac{4d^2}{D^2})^{-3/2} ight]$.Using the binomial approximation $(1+x)^n \approx 1+nx$ for $x $E \approx \frac{2kq}{D^2} \left[ (1 - \frac{3d^2}{2D^2}) - (1 - \frac{6d^2}{D^2}) ight]$.$E \approx \frac{2kq}{D^2} \left[ \frac{6d^2}{D^2} - \frac{3d^2}{2D^2} ight] = \frac{2kq}{D^2} \left[ \frac{9d^2}{2D^2} ight] = \frac{9kqd^2}{D^4}$.Thus,$E \propto \frac{1}{D^4}$.

Four point charges $-q, +q, +q$ and $-q$ are placed on the $y$-axis at $y = -2d, y = -d, y = +d$ and $y = +2d$,respectively. The magnitude of the electric field $E$ at a point on the $x$-axis at $x = D$,with $D >> d$,will vary as:

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