Potential Due to a Point Charge — Question 4

Question 4

A finite reference sphere moves the zero but leaves the radial work unchanged

A point charge Q produces a potential whose zero is assigned at radius R rather than infinity.

1. Derive V(r) and identify where it is positive and negative for Q>0.

2. Find the quasistatic work moving q from R/2 to 2R and verify reference independence.

Question 4 - Solution

A finite reference sphere moves the zero but leaves the radial work unchanged

(a) Method and calculation

$$V(r)=\frac{kQ}{r}-\frac{kQ}{R}=kQ\left(\frac1r-\frac1R\right).$$$$Q>0:\quad V>0\ (r<R),\quad V=0\ (r=R),\quad V<0\ (r>R).$$

(b) Interpretation and further calculation

$$W_{\mathrm{ext}}=q[V(2R)-V(R/2)]=-\frac{3kQq}{2R}.$$$$q\left[\frac{kQ}{2R}-\frac{kQ}{R/2}\right]=-\frac{3kQq}{2R}.$$

Physical check

A positive charge can produce negative assigned V outside the reference sphere; only the infinity-reference sign directly tracks Q.