Question

Calculate the potential at the center of a square of side √2 m, which carries at its 4 corner charges of

2nC , 1nC , -2nC , -3nC respectively. pls someone give me answer asap​

Answer 1

Answer:

v = -18 C/m

Explanation:

First of all Side of the square =$\sqrt{2} m$

So, Length of the diagonal of the square = $\sqrt{(\sqrt{2})^2+ (\sqrt{2})^2 }$= 2 m

Now, the 2 diagonals of a square bisect each other exactly at its  center.

Therefore, distance of Center(O) from ecah vertex = 2/2 = 1 m

Now the potential due to a point charge is given by -

V=kQ/r       where k is a constant equal to 9.0×109 N⋅m2/C2 .

⇒ V at Center of square (O) = K$(\frac{q_1}{r_1} + \frac{q_2}{r_2} + \frac{q_3}{r_3} + \frac{q_4}{r_4} )$

or V = 9 ×$10^{9}$(2/1 + 1/1 + (-2/1) + (-3/1)×$10^{-9}$

or V=  9 (2+1 - 2 - 3) =  -18 C/m

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