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: Understanding the Concept
Potential is a scalar, so the potentials of the eight charges simply add. Every corner is at the same distance from the centre.

Step 2: Distance
The body diagonal of a cube of side \(r\) is \(r\sqrt3\). The centre is halfway along it:
\[ d=\frac{r\sqrt3}{2} \]

Step 3: Potential of one charge
\[ V_1=\frac{Q}{4\pi\varepsilon_0d}=\frac{Q}{4\pi\varepsilon_0}\cdot\frac{2}{r\sqrt3} \]

Step 4: Total
\[ V=8V_1=\frac{8Q}{4\pi\varepsilon_0}\cdot\frac{2}{r\sqrt3}=\frac{4Q}{\pi\varepsilon_0r\sqrt3} \]

Step 5: Conclusion
This is option (A). Option (B) doubles the result, which happens if the distance is taken as \(r\sqrt3/4\).

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