1. Given data

• Charge at origin: $q = 10^{-6}\,\mu C = 10^{-6}\times 10^{-6}C = 10^{-12}C$
• Point $P(\sqrt{3},\sqrt{3})$
• Point $Q(\sqrt{6},0)$

2. Find distances of $P$ and $Q$ from origin

For point $P$: $OP = \sqrt{(\sqrt{3})^2+(\sqrt{3})^2} = \sqrt{3+3} = \sqrt{6}\,m$

For point $Q$: $OQ = \sqrt{(\sqrt{6})^2+0^2} = \sqrt{6}\,m$

So both points are at the same distance from the charge.

3. Use formula for electric potential due to a point charge

Potential at distance $r$ from a point charge is $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$

Since both $P$ and $Q$ have the same distance $\sqrt{6}$ from the origin, $V_P = V_Q$

Therefore, the potential difference is $V_P - V_Q = 0$

4. Check options

• A: $\sqrt{3}\,V$ ❌
• B: $\sqrt{6}\,V$ ❌
• C: $0\,V$ ✅
• D: $3\,V$ ❌

Hence the correct option is C.