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(A) Given that,the point charges are placed on the vertices of a square of side $a = 2.8 \,m$. Let $r$ be the distance of the centre $O$ from each cor

Vedclass · Accepted Answer

$190.89$

Vedclass · Answer

(A) Given that,the point charges are placed on the vertices of a square of side $a = 2.8 \,m$. Let $r$ be the distance of the centre $O$ from each corner.The diagonal of the square is $d = \sqrt{a^2 + a^2} = a\sqrt{2}$.The distance $r$ from the centre to each corner is half of the diagonal:$r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}} = \frac{2.8}{\sqrt{2}} \,m$.The electric potential $V$ at a point due to a point charge $q$ is given by $V = \frac{Kq}{r}$,where $K = 9 	imes 10^9 \,N \cdot m^2/C^2$.The total electric potential $V_0$ at the centre $O$ due to all four charges $q_1, q_2, q_3,$ and $q_4$ is:$V_0 = V_1 + V_2 + V_3 + V_4 = \frac{K}{r}(q_1 + q_2 + q_3 + q_4)$.Given charges: $q_1 = +20 \,nC$,$q_2 = +40 \,nC$,$q_3 = -34 \,nC$,$q_4 = +16 \,nC$.Sum of charges: $\sum q = (20 + 40 - 34 + 16) \,nC = 42 \,nC = 42 	imes 10^{-9} \,C$.Substituting the values:$V_0 = \frac{9 	imes 10^9 	imes 42 	imes 10^{-9}}{2.8 / \sqrt{2}} = \frac{9 	imes 42 	imes \sqrt{2}}{2.8} = \frac{378 	imes 1.414}{2.8} = 135 	imes 1.414 = 190.89 \,V$.

Four point charges are placed at the vertices of a square of side $2.8 \,m$ as shown in the figure. Find the electric potential at the centre of the square. (in $\,V$)

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