Potential Due to a Group of Point Charges — Question 7

Question 7

Moving between two observation points needs only a scalar voltage sum

Use $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$. Fixed charges +3.00 nC and -2.00 nC are .500 m apart. A lies .300 m from each source; B lies .500 m from each. Use zero at infinity.

1. Find VA and VB, using the geometry to justify the equal-distance sums.

2. Find the change in energy of a $-4.00$ nC probe transferred from A to B.

Question 7 - Solution

Moving between two observation points needs only a scalar voltage sum

(a) Method and calculation

$$V_A=k\frac{(3.00-2.00)\times10^{-9}}{.300}=29.9667\,\mathrm V.$$$$V_B=k\frac{1.00\times10^{-9}}{.500}=17.9800\,\mathrm V.$$

(b) Interpretation and further calculation

$$\Delta V=-11.9867\,\mathrm V.$$$$\Delta U=(-4.00\,\mathrm{nC})(-11.9867\,\mathrm V)=+47.9467\,\mathrm{nJ}.$$

Physical check

Both distances exceed half the source separation, so both observation points exist on the perpendicular bisector.