1. Potential at a point is the work done required bring a unit charge from infinity to that point without acceleration.
2. Potential due To A point charge:
Let P be a point at a distance Y from a charge +q. Let A be a point at a distance ‘x’ from q, and E is directed along PA. Consider a positive charge at A. Then the electric field intensity at ‘A’ is given by
E = \(\frac{1}{4\pi\varepsilon_0}\) \(\frac{q}{x^2}\)
If this unit charge is moved (opposite to E} through a distance dx, the work done dw = – Edx.
[-ve sign indicates that dx is opposite to E ] So the potential at ‘P’ is given by
V = \(\frac{+q_1}{4\pi\varepsilon_0r}\) (since \(\frac{1}{\infty}\)= 0).
3. V = \(\frac{1}{4\pi\varepsilon_0}\) \(\frac{q}{R}\) = \(\frac{9\times10^9\times250\times10^{-5}}{10\times10^{-2}}\)
= 2.25 × 107v.