Question

Three charges $ 1\mu C, $ $ 1\mu C $ and $ 2\mu C $ are kept at vertices of $A, B$ and $C$ of an equilateral triangle $ABC$ of $ 10\,cm $ side respectively. The resultant force on the charge at $C$ is

• A. 0.9 N
• B. 1.8 N
• C. 2.72 N
• D. 3.12 N

Answer 1

Answer: (D)

The situation is shown in figure. Force on $C$ due to $A$

$F_{A C} =\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{A} q_{C}}{r_{A C}^{2}} $
$=\frac{9 \times 10^{9} \times 1 \times 2 \times 10^{-12}}{(0.1)^{2}}=1.8 \,N $
Similarly, $ F_{B C}=1.8\, N $
Hence, net force on
$C=F_{A C}+F_{B C}=3.6\, N$