Question

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?

Answer 1

The electric potential at the center of the square due to all four charges is 4V₀, and the magnitude of the electric field at the center is 4E₀.

Step-by-step explanation:

To find the electric potential and magnitude of the electric field at the center of the square due to all four charges, we can use the principle of superposition. The electric potential at the center of the square is the sum of the electric potentials due to each individual charge. In this case, since all four charges are at the corners of the square, the electric potential at the center due to one charge alone, V₀, is the same for all charges. Therefore, the total electric potential at the center is 4V₀.

The magnitude of the electric field at the center of the square is also the sum of the electric fields due to each individual charge. Since all charges are positive and have the same magnitude, the electric fields will point away from the charges and towards the center of the square. The magnitude of the electric field due to one charge alone, E₀, is the same for all charges. Therefore, the total magnitude of the electric field at the center is 4E₀.

Answer 2

Step-by-step 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.

`V_(o)=(Kq)/(d)`

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