Question

Three charges of each magnitude $100\mu C$ are placed at the corners $A,B$ and $C$ of an equilateral triangle of side $4m$. If the charges at points $A$ and $C$ are positive and the charge at point $B$ is negative, then the magnitude of total force acting on the charge at $C$ and angle made by it with $AC$ are

• A. $5.625N,60^{\circ }$
• B. $0.5625N,60^{\circ }$
• C. $5.625N,30^{\circ }$
• D. $0.5625N,30^{\circ }$

Answer 1

Answer: (A)

According to the question, we can draw the following diagram,

From the question, it clear that the charge on the each corner is $100\mu C$.
So, $F_{net}=F$
$=\frac{k{Q}_1{Q}_2}{r^2}\dots (i)$
Given, $Q_1=Q_2=100\mu C=100\times 10^{-6}C$,
$\left[1\mu C=10^{-6}C\right]$
and $r=4m$
Putting the given values in Eq. (i), we get
$=\frac{9\times 10^9\times {\left(100\times 10^{-6}\right)}^2}{(4{)}^2}$
$F_{\text{net }}=5.625N$ ,
$60^{\circ }$