Q121 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
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)
- Correct answerA$\dfrac{4Q}{\pi\epsilon_0 r\sqrt{3}}$
- B$\dfrac{8Q}{\pi\epsilon_0 r\sqrt{3}}$
- C$\dfrac{16Q}{3\pi\epsilon_0 r}$
- D$\dfrac{32Q}{\pi\epsilon_0 r\sqrt{3}}$
Solution
The length of the body diagonal of a cube of side $r$ is $r\sqrt{3}$.
The distance from the centre of the cube to any corner is half of the body diagonal, which is $d = \dfrac{r\sqrt{3}}{2}$.
The electric potential at the centre due to a single charge $Q$ at one corner is $V_1 = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{d}$.
Since there are $8$ identical charges at the $8$ corners, the total potential at the centre is $V = 8 V_1$.
Substituting the value of $d$:
$V = 8 \times \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{\dfrac{r\sqrt{3}}{2}}$
$V = \dfrac{16Q}{4\pi\epsilon_0 r\sqrt{3}}$
$V = \dfrac{4Q}{\pi\epsilon_0 r\sqrt{3}}$
Answer: $\dfrac{4Q}{\pi\epsilon_0 r\sqrt{3}}$
The distance from the centre of the cube to any corner is half of the body diagonal, which is $d = \dfrac{r\sqrt{3}}{2}$.
The electric potential at the centre due to a single charge $Q$ at one corner is $V_1 = \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{d}$.
Since there are $8$ identical charges at the $8$ corners, the total potential at the centre is $V = 8 V_1$.
Substituting the value of $d$:
$V = 8 \times \dfrac{1}{4\pi\epsilon_0} \dfrac{Q}{\dfrac{r\sqrt{3}}{2}}$
$V = \dfrac{16Q}{4\pi\epsilon_0 r\sqrt{3}}$
$V = \dfrac{4Q}{\pi\epsilon_0 r\sqrt{3}}$
Answer: $\dfrac{4Q}{\pi\epsilon_0 r\sqrt{3}}$
Hence, the correct answer is option: 'A'.