Answer:
Explanation:
The final net force will be in the Z- direction. Let's find out the z component of the force on the differential volume of charge is:
df = dqEcosθz
[tex]E = \frac{1}{4\pi epsilon} \frac{Qr}{R^{3} }[/tex]
dq = ρdV = [tex]\frac{3Q}{4\pi R^{3} }[/tex][tex]r^{2}[/tex]dr.sinθdθdΦ
integrate it over half ball,
[tex]F_{z} = \int\limits^._V {df_{x}dV} =\frac{1}{4\pi epsilon } \frac{Q}{R^{3} } \frac{3Q}{4\pi R^{3} }\int\limits^R_0 {\int\limits^\frac{\pi }{2} _{0} {\int\limits^\frac{\pi }{2} _0 {r^{3} } \, dr } \, } \,[/tex].sinθcosθdθdΦ.( these are part of the integral, i was unable to write it in equation format).
= [tex]\frac{3Q^{2} }{32\pi epsilonR^{2} } \int\limits^\frac{\pi }{2} _b {} \,[/tex] sinθcosθdθ
= [tex]\frac{3Q^{2} }{64\pi epsilon R^{2} }[/tex]
[tex]F = \frac{3Q^{2} }{64\pi epsilon R^{2} } z[/tex]
