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Four positive charges of magnitude q are ar- ranged at the corners of a square, as shown. At the center C of the square, the potential due to one charge alone is V0, and the electric field due to one charge alone has magnitude E0. Which of the following correctly gives the electric potential and the magnitude of the electric field at the center of the square due to all four charges?

Respuesta :

Answer:

Explanation:

Let the four charges is q and the side of square is a.

Let the distance of all charges from the centre is d.

AO = BO = CO = DO = d

Potential due to single charge at the centre is Vo.

[tex]V_{o}=\frac{Kq}{d}[/tex]

Potential at the centre due to all charges is

V = Vo + Vo + Vo + Vo = 4 Vo

Electric field due to single charge at centre is Eo.

So, the electric ifled due to all charges at centre is

E = (Eo - Eo) + (Eo - Eo) = 0

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