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Four charges of magnitude +q are placed at the corners of a square whose sides have a length d. What is the magnitude of the total force exerted by the four charges on a charge Q located a distance b along a line perpendicular to the plane of the square and equidistant from the four charges?

Respuesta :

Answer:

[tex]F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

Explanation:

Since all the four charges are equidistant from the position of Q

so here we can assume this charge distribution to be uniform same as that of a ring

so here electric field due to ring on its axis is given as

[tex]E = \frac{k(4q)x}{(x^2 + R^2)^{1.5}}[/tex]

here we have

x = b

and the radius of equivalent ring is given as the distance of each corner to the center of square

[tex]R = \frac{d}{\sqrt2}[/tex]

now we have

[tex]E = \frac{4kq b}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

so the force on the charge is given as

[tex]F = QE[/tex]

[tex]F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}[/tex]

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