Question

A charge \(Q\) is placed at each corner of a cube of side \(r\). The potential at the centre of the cube is (\(ε_0\) = permittivity of free space)

• A. \(\frac{4Q}{πε_0r\sqrt{3}}\)
• B. \(\frac{8Q}{πε_0r\sqrt{3}}\)
• C. \(\frac{16Q}{3πε_0r}\)
• D. \(\frac{32Q}{πε_0r\sqrt{3}}\)

Hint:

All eight charges are at the same distance from the centre, so add their potentials.

Answer 1

The Correct Option is A

Step 1: Approach
Use the single-charge potential formula with the distance written as $d$ and substitute at the end.

Step 2: Formula
$V=8\cdot\dfrac{kQ}{d}$ with $k=\dfrac1{4\pi\varepsilon_0}$. The distance from a corner to the centre is half the space diagonal, $d=\dfrac{\sqrt3\,r}{2}$.

Step 3: Compute
$V=\dfrac{8kQ\cdot2}{\sqrt3\,r}=\dfrac{16kQ}{\sqrt3\,r}=\dfrac{16Q}{4\pi\varepsilon_0\sqrt3\,r}=\dfrac{4Q}{\pi\varepsilon_0\sqrt3\,r}$.

Step 4: Answer
Option (A).

Final Answer:
Eight equal potentials at distance r root 3 over 2 give 4Q over pi epsilon-0 r root 3, option (A). \[ \boxed{\frac{4Q}{\pi\varepsilon_0r\sqrt3}} \]