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ABC is an equilateral triangle of each side $5 \;cm$ . Two charges of $\pm \large\frac{50}{3}$$\times 10^{-3} \mu c$ are placed at A and B respectively . Calculate the magnitude and direction of resultant intensity at C.

$\begin{array}{1 1} 6 \times 10^4 \;NC^{-1} \\ 6 \times 10^{-4} \;NC^{-1} \\ 6 \times 10^3 \;NC^{-1} \\ 6 \times 10^{-3} \;NC^{-1} \end{array} $

1 Answer

Solution :
The Point c lies on the equatorial line of electric dipole formed by charge $\pm \large\frac{50}{3} $$ \times 10^{-3} \mu c$ placed at A and B respectively
Hence field intensity due to the change $q_2$
$\large\frac{1}{4 \pi \in_0}$ along BC
$\large\frac{9 \times 10^9 \times \Large\frac{50}{3} \times 10^3 \times 10^{-6}}{25 \times 10^{-4}}$
$\qquad= 6 \times 10^{4}\;NC^{-1}$
Answer : $ 6 \times 10^4 \;NC^{-1}$
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