Question

Two charges of $+200\mu C$ and $-200\mu C$ are placed at the corners B and C of an equilateral triangle ABC of side $0.1m$. The force on a charge of $5\mu C$ placed at A is :

• A. $1800N$
• B. $1200\sqrt{3}N$
• C. $900N$
• D. $600\sqrt{3}N$

Answer 1

Answer: (C)

$\to$ The vertical components of force cancel out as the magnitude of charges are equal.
Net force experience by A is
$F_{net}=\frac{k(200\times {10}^{-6})(5\times {10}^{-6})}{(0\cdot 1{)}^2}\times Cos\left({60}^0\right)+\frac{k(200\times {10}^{-6})(5\times {10}^{-6})}{(0\cdot 1{)}^2}\left(Cos60\right)$

$=\frac{(9\times {10}^9)(200\times {10}^{-6})(5\times {10}^{-6})}{{10}^{-2}}$

$=9\times 20\times 5$
$F_{net}=900N\left(\hat{x}\right)$