Applying Gauss' Law: Spherical Symmetry — Question 2

Question 2

Treating an interior point as outside a charged sphere can overestimate its field

A solid nonconducting sphere of radius R carries uniform density $\rho>0$ and total charge Q. There are no other sources.

1. Derive the field inside and outside in terms of Q and R.

2. At $r=R/2$, compare the actual field with the incorrect point-charge formula using the entire Q at the center. Quantify the overestimate.

Question 2 - Solution

Treating an interior point as outside a charged sphere can overestimate its field

(a) Method and calculation

$$Q_{\mathrm{enc}}(r)=\frac43\pi\rho r^3=Q(r/R)^3\quad(r<R).$$$$E_r=\frac{Qr}{4\pi\epsilon_0R^3}\quad(r<R),\qquad E_r=\frac Q{4\pi\epsilon_0r^2}\quad(r>R).$$

(b) Interpretation and further calculation

$$\frac{E_{\mathrm{actual}}(r)}{E_{\mathrm{point}}(r)}=(r/R)^3\Rightarrow\frac{E_{\mathrm{actual}}(R/2)}{E_{\mathrm{point}}(R/2)}=\frac18.$$$$\frac{E_{\mathrm{point}}-E_{\mathrm{actual}}}{E_{\mathrm{actual}}}=7=700\%.$$Only the charge enclosed by the interior Gaussian sphere contributes to its radial field after symmetry is used.

Physical check

The fields agree at R and the interior field vanishes linearly at the center.