Three points charges are placed at the corner | vedclass

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Question

For equilateral triangle $\mathrm{AOB}, A O=O B=B A=a$ and $A C=O C=B C=r$

The potential at centroid $\mathrm{C}$ is $V=k \frac{2 q}{A C}+k \frac{-

Vedclass · Accepted Answer

$A$ and $B$ both

Vedclass · Answer

For equilateral triangle $\mathrm{AOB}, A O=O B=B A=a$ and $A C=O C=B C=r$

The potential at centroid $\mathrm{C}$ is $V=k \frac{2 q}{A C}+k \frac{-q}{O C}+k \frac{-q}{B C}=k \frac{2 q}{r}-k \frac{q}{r}-k \frac{q}{r}=0$

The net electric field at centroid $\mathrm{C}$ is $E=k \frac{2 q}{r^{2}}-k \frac{q}{r^{2}} \cos 30-k \frac{q}{r^{2}} \cos 30=k(0.2) \frac{2 q}{r^{2}} 
eq$ $0$

As the total charge is zero so for dipole moment the choice of origin is independent. we assume $O$ as origin.

dipole moment, $p=-q(0)-q(L \hat{i})+2 q\left(\frac{L}{2} \hat{i}+\frac{\sqrt{3} L}{2} \hat{j}ight)$

or $p=\sqrt{3} q L \hat{j}$

Three points charges are placed at the corners of an equilateral triangle of side $L$ as shown in the figure.

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