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Question

(A) The total potential energy $U_{total}$ of the system is the sum of the potential energy of the dipole (self-energy) and the interaction energy of

Vedclass · Accepted Answer

$\frac{1}{{4\pi {\varepsilon _0}}}\left[ { - \frac{{{q^2}}}{d} - \frac{{qQd}}{{{D^2}}}} \right]$

Vedclass · Answer

(A) The total potential energy $U_{total}$ of the system is the sum of the potential energy of the dipole (self-energy) and the interaction energy of the dipole with the charge $Q$.$1$. The self-energy of the dipole (charges $+q$ and $-q$ separated by $d$) is $U_{dipole} = -\frac{1}{4\pi \varepsilon_0} \frac{q^2}{d}$.$2$. The interaction energy of the dipole with charge $Q$ is $U_{interaction} = V_Q(+q) + V_Q(-q) = \frac{1}{4\pi \varepsilon_0} Q \left( \frac{q}{D} - \frac{q}{D+d} \right)$.$3$. Since $D >> d$,we can use the binomial approximation: $\frac{1}{D+d} = \frac{1}{D(1 + d/D)} \approx \frac{1}{D} (1 - d/D) = \frac{1}{D} - \frac{d}{D^2}$.$4$. Substituting this into the interaction energy: $U_{interaction} \approx \frac{1}{4\pi \varepsilon_0} Q \left( \frac{q}{D} - q(\frac{1}{D} - \frac{d}{D^2}) \right) = \frac{1}{4\pi \varepsilon_0} \frac{qQd}{D^2}$.Wait,looking at the figure,the distance from $Q$ to $-q$ is $D$,and from $Q$ to $+q$ is $D+d$. Thus,$U_{interaction} = \frac{1}{4\pi \varepsilon_0} Q \left( \frac{-q}{D} + \frac{q}{D+d} \right) = \frac{qQ}{4\pi \varepsilon_0} \left( \frac{1}{D+d} - \frac{1}{D} \right) \approx \frac{qQ}{4\pi \varepsilon_0} \left( \frac{1}{D} - \frac{d}{D^2} - \frac{1}{D} \right) = -\frac{qQd}{4\pi \varepsilon_0 D^2}$.Therefore,$U_{total} = -\frac{1}{4\pi \varepsilon_0} \frac{q^2}{d} - \frac{1}{4\pi \varepsilon_0} \frac{qQd}{D^2} = \frac{1}{4\pi \varepsilon_0} \left[ -\frac{q^2}{d} - \frac{qQd}{D^2} \right]$.

$A$ system of three charges is placed as shown in the figure. If $D >> d$,the potential energy of the system is best given by:

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