Elaborate marked step.

Elaborate marked step. voo LTE 56%" 18:57 Page 122 v Solution to Q.31 sin Total vertical component of force, F 2Fsin9 4neo 2TIED qQx 3/2 21180 dFI _ um force, dx -3/2 Exit

Dear Student,

F' = \frac{1}{2 {πε}_{0}} \times \frac{qQx}{{[{(\frac{d}{2})}^{2} + x^{2}]}^{\frac{3}{2}}} F' = \frac{qQ}{2 {πε}_{0}} \times x {[{(\frac{d}{2})}^{2} + x^{2}]}^{\frac{- 3}{2}} . . . . . . . (1) for maximum force, \frac{dF'}{dx} = 0 we know that, differenciation of u . v = v \frac{du}{dx} + u \frac{dv}{dx} 0 = \frac{qQ}{2 {πε}_{0}} \times [{\{{(\frac{d}{2})}^{2} + x^{2}\}}^{\frac{- 3}{2}} \times 1 + x \times (- \frac{3}{2}) {\{{(\frac{d}{2})}^{2} + x^{2}\}}^{- \frac{3}{2} - 1} \times 2 x] 0 = \frac{qQ}{2 {πε}_{0}} \times [{\{{(\frac{d}{2})}^{2} + x^{2}\}}^{- \frac{3}{2}} - x^{2} \times 3 {\{{(\frac{d}{2})}^{2} + x^{2}\}}^{- \frac{5}{2}}] 0 = {\{{(\frac{d}{2})}^{2} + x^{2}\}}^{- \frac{5}{2}} \times [{\{{(\frac{d}{2})}^{2} + x^{2}\}}^{- 1} - 3 x^{2}] 0 = [\{{(\frac{d}{2})}^{2} + x^{2}\} - 3 x^{2}] 0 = {(\frac{d}{2})}^{2} + x^{2} - 3 x^{2}