0
$\begingroup$

From Griffiths, intro to electrodyanmics:

enter image description here

Now, I approach the problem a different way then he does, and I miss the answer by a factor of half

The way I do it is that I find the electric field outside the metal sphere using Gauss' law which comes up to be $E$=$\frac{1}{4\pi\epsilon_0}$ $\frac{Q_{enc}}{r^2}$

Then the Electric field on the surface of the sphere is $E$=$\frac{1}{4\pi\epsilon_0}$ $\frac{Q_{enc}}{R^2}$

I divide that by 2 to get the contribution of only a hemisphere (let that be the southern hemisphere) to the electric field which is $E$=$\frac{1}{8\pi\epsilon_0}$ $\frac{Q_{enc}}{r^2}$

Now the formula for electrostatic pressure is $P$=$\frac{\epsilon_0}{2} E^2$ , where E is the electric field just outside the sphere (same as electric field of the hemisphere on the surface), then:

$P$= $\frac{\epsilon_0}{2} \frac{1}{64\pi^2\epsilon_0^2}$ $\frac{Q^2_{enc}}{R^4}$ = $\frac{1}{128\pi^2\epsilon_0}$$\frac{Q^2_{enc}}{R^4}$

Now, the problem is when I want to find the force due to that pressure I have to multiply that pressure by only half the surface area of the sphere which is the surface area of the hemisphere I'm taking the pressure at

$F = PA$

where A = $2\pi R^2$ then,

$F=$$\frac{1}{128\pi^2\epsilon_0}$$\frac{Q^2_{enc}}{R^4}$ $2\pi R^2$

I end up with $F = $$\frac{1}{64\pi\epsilon_0}$$\frac{Q^2_{enc}}{R^2}$

which should be $1/32$ instead of $1/64$ , where have I gone wrong please, Griffiths did it by integrating the force per unit area on every surface element on the northern hemisphere, but I believe I could do it without integration, where have I gone wrong please?

$\endgroup$
1
$\begingroup$

You went wrong when you divided the field of the entire sphere by two to get the field due to one hemisphere. The field of a half-sphere is nothing like the field of a sphere. It isn’t even radial, because there is no spherical symmetry.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see, does that mean I'm forced to do it by integration and getting the average field between inside and outside? $\endgroup$ – khaled014z Nov 26 '18 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.