Question 4.32: Earnshaw’s theorem (Prob. 3.2) says that you cannot trap a c...

(a) According to Eqs. 4.1 and 4.5,  $F =\alpha( E \cdot \nabla) E$ . From the product rule,

$p =\alpha E$                (4.1)

$F =( p \cdot \nabla ) E$                   (4.5)

$\nabla E^{2}=\nabla( E \cdot E )=2 E \times(\nabla \times E )+2( E \cdot \nabla) E$ .

But in electrostatics  $\nabla \times E = 0 , \text { so }( E \cdot \nabla) E =\frac{1}{2} \nabla\left(E^{2}\right)$ , and hence

$F =\frac{1}{2} \alpha \nabla\left(E^{2}\right)$ .

[It is tempting to start with Eq. 4.6, and write  $F =-\nabla U=\nabla( p \cdot E )=\alpha \nabla( E \cdot E )=\alpha \nabla\left(E^{2}\right)$ . Theerror occurs in the third step: p should not have been differentiated, but after it is replaced by $\alpha E$ we aredifferentiating both E’s.]

$U=- p \cdot E$                         (4.6)

 (b) Suppose $E^{2}$ has a local maximum at point P . Then there is a sphere (of radius R) about P such that $E^{2}\left(P^{\prime}\right)<E^{2}(P)$ , and hence  $\left| E \left(P^{\prime}\right)\right|<| E (P)|$ |, for all points on the surface. But if there is no charge inside the sphere,  then Problem 3.4a says the average field over the spherical surface is equal to the value at the center:

$\frac{1}{4 \pi R^{2}} \int E d a= E (P)$ .

or, choosing the z axis to lie along E(P),

$\frac{1}{4 \pi R^{2}} \int E_{z} d a=E(P)$ .

But if $E^{2}$ has a maximum at P, then

$\int E_{z} d a \leq \int| E | d a<\int| E (P)| d a=4 \pi R^{2} E(P)$ ,

and it follows that  $E(P)<E(P)$ , a contradiction. Therefore,   $E^{2}$ cannot have a maximum in a charge-free region. [It can have a minimum, however; at the midpoint between two equal charges the field is zero, and this is obviously a minimum.]