If OP = 1 , cm and OS = 2 , cm,calculate the | vedclass

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(A) The electric potential due to a dipole at a point $(r, 	heta)$ is given by $V = \frac{kp \cos 	heta}{r^2}$,where $k = 9 	imes 10^9 \, N \cdot m

Vedclass · Accepted Answer

$\frac{100}{3} \, J$

Vedclass · Answer

(A) The electric potential due to a dipole at a point $(r, 	heta)$ is given by $V = \frac{kp \cos 	heta}{r^2}$,where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.For point $P$: $r_P = 1 	imes 10^{-2} \, m$,$	heta_P = 45^{\circ}$.$V_P = \frac{kp \cos 45^{\circ}}{r_P^2} = \frac{(9 	imes 10^9)(2 	imes 10^{-6})(\frac{1}{\sqrt{2}})}{(10^{-2})^2} = \frac{18 	imes 10^3}{\sqrt{2} 	imes 10^{-4}} = \frac{18 	imes 10^7}{\sqrt{2}} \, V$.For point $S$: $r_S = 2 	imes 10^{-2} \, m$,$	heta_S = 135^{\circ}$.$V_S = \frac{kp \cos 135^{\circ}}{r_S^2} = \frac{(9 	imes 10^9)(2 	imes 10^{-6})(-\frac{1}{\sqrt{2}})}{(2 	imes 10^{-2})^2} = \frac{-18 	imes 10^3}{\sqrt{2} 	imes 4 	imes 10^{-4}} = \frac{-18 	imes 10^7}{4\sqrt{2}} \, V$.The work done by the electric field is $W = -q(V_S - V_P) = q(V_P - V_S)$.$W = (\frac{4\sqrt{2}}{27} 	imes 10^{-6}) \left[ \frac{18 	imes 10^7}{\sqrt{2}} - (-\frac{18 	imes 10^7}{4\sqrt{2}}) ight] = (\frac{4\sqrt{2}}{27} 	imes 10^{-6}) \left[ \frac{18 	imes 10^7}{\sqrt{2}} (1 + \frac{1}{4}) ight]$.$W = (\frac{4\sqrt{2}}{27} 	imes 10^{-6}) \left[ \frac{18 	imes 10^7}{\sqrt{2}} 	imes \frac{5}{4} ight] = \frac{4\sqrt{2}}{27} 	imes \frac{18 	imes 10^1 	imes 5}{4\sqrt{2}} = \frac{180 	imes 5}{27} = \frac{900}{27} = \frac{100}{3} \, J$.

If $OP = 1 \, cm$ and $OS = 2 \, cm$,calculate the work done by the electric field in shifting a point charge $q = \frac{4\sqrt{2}}{27} \, \mu C$ from point $P$ to $S$ in the given figure. The dipole moment is $p = 2 \times 10^{-6} \, C \cdot m$.

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