A charge Q is placed at each corner of a cube | vedclass

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Question

(A) cube has $8$ corners. Let the side length of the cube be $r$.The distance from any corner of the cube to its centre is half of the body diagonal.T

Vedclass · Accepted Answer

$\frac{4Q}{\pi\epsilon_0r\sqrt{3}}$

Vedclass · Answer

(A) cube has $8$ corners. Let the side length of the cube be $r$.The distance from any corner of the cube to its centre is half of the body diagonal.The body diagonal of a cube with side $r$ is $d = r\sqrt{3}$.Therefore, the distance $a$ from each corner to the centre is $a = \frac{r\sqrt{3}}{2}$.The electric potential $V$ due to a point charge $Q$ at a distance $a$ is given by $V = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{a}$.Since there are $8$ identical charges at the corners, the total potential at the centre is the sum of the potentials due to each charge:$V_{total} = 8 	imes \left( \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{a} ight) = \frac{8Q}{4\pi\epsilon_0 \cdot (\frac{r\sqrt{3}}{2})} = \frac{8Q}{2\pi\epsilon_0 r\sqrt{3}} = \frac{4Q}{\pi\epsilon_0 r\sqrt{3}}$.Thus, the correct option is $A$.

$A$ charge $Q$ is placed at each corner of a cube of side $r$. The potential at the centre of the cube is ($\epsilon_0$ = permittivity of free space).

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