From Griffiths, intro to electrodyanmics:

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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?


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.

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  • $\begingroup$ 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

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