(4-70) ER(R, 0) = av(R, 0) ƏR Using Eq. (4-67) in Ed (4-70) and noting Eas. (4-68) and (4-69), wo haver Logo D - Q R a a[R-a²/d) cos 0] cos A ER(R, 0) = 4ñ€。 ((R² + d² − 2Rd cos 0)³/² — d[R² + (a²/d)² - 2R(a²/d) cos 01³/² (4-71) a) To find the induced surface charge on the sphere, we set R = a in Eq. (4-71) and evaluate Ps= €0ER(a,0), (4-72) which yields the following after simplification: Q(d² - a²) Ps= 4na(a² + d²2ad cos 0)3/2 (4-73) Ea (4-73) tells us that the induced surface charge in
where, by the law of cosines, and of 12₁ Q 1 V(R, 0) = 4TEO Ro dRo R₂ = [R² + d² - 2Rd cos 0]¹/2 2 71/2 R₂₁ - [R² + (G)" - 2R (G) cs 0]" (2 Ro AP(R, 0) FIGURE 4-12 R Ro RQ₁ high Jong Q (4-67) (4-68) (4-69) mputing induced charge distribution
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