Four charges 2C, -3C, -4C and 5C respectively | vedclass

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(B) Let the side of the square be $L$. The distance of each corner from the center $O$ is $r = \frac{L}{\sqrt{2}}$.The electric potential at the cente

Vedclass · Accepted Answer

Electric field is non-zero but electric potential is zero.

Vedclass · Answer

(B) Let the side of the square be $L$. The distance of each corner from the center $O$ is $r = \frac{L}{\sqrt{2}}$.The electric potential at the center $O$ is the algebraic sum of potentials due to individual charges:$V = \frac{k}{r} (q_1 + q_2 + q_3 + q_4) = \frac{k}{r} (2 - 3 - 4 + 5) = \frac{k}{r} (0) = 0$.The electric field at the center is the vector sum of fields due to individual charges. The field due to a pair of opposite charges $q_i$ and $q_j$ at the center is $\vec{E}_{ij} = \frac{k(q_i - q_j)}{r^2} \hat{r}$.For the pair $(2C, -4C)$,the net field is directed towards $-4C$ with magnitude $\frac{k(2 - (-4))}{r^2} = \frac{6k}{r^2}$.For the pair $(-3C, 5C)$,the net field is directed towards $5C$ with magnitude $\frac{k(5 - (-3))}{r^2} = \frac{8k}{r^2}$.Since these two resultant vectors are perpendicular,the total electric field is $E = \sqrt{(\frac{6k}{r^2})^2 + (\frac{8k}{r^2})^2} = \frac{k}{r^2} \sqrt{36 + 64} = \frac{10k}{r^2} = \frac{10k}{(L/\sqrt{2})^2} = \frac{20k}{L^2} 
eq 0$.

Four charges $2C, -3C, -4C$ and $5C$ respectively are placed at the corners of a square of side length $L$. Which of the following statements is true for the point of intersection of the diagonals?

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