# Electrostatic force per unit area on a hemisphere due to its other half

From Griffiths, intro to electrodyanmics:

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?