Question

Charges $5\times {10}^{-7}\text{C},-2.5\times {10}^{-7}\text{C}$ and $1\times {10}^{-7}\text{C}$ are held fixed at the three corners $A,B$ and $C$ of an equilateral triangle of side $10\text{cm}$ respectively. Find the electric force on the charge at $C$ due to the rest two
(correct answer + 1, wrong answer - 0.25)

• A. $0.0025\text{N}$
• B. $0.038\text{N}$
• C. $0.040\text{N}$
• D. $0.045\text{N}$

Answer 1

The correct option is B $0.038\text{N}$

Side of triangle $=AB=BC=CA=10\text{cm}=0.1\text{m}$
From Coulomb's law force between two charges separated by a distance is given as
$F=\frac{k{q}_1{q}_2}{r^2}$
Here $k$ is a constant whose value is $9\times {10}^9{\text{Nm}}^2{\text{/C}}^2$ and $r=0.1\text{m}$
We have shown the direction of force between charges at $A$ & $C$ and $B$ & $C$
For charges at $A$ and $C$:
$F_1=\frac{9\times {10}^9\times 5\times {10}^{-7}\times 1\times {10}^{-7}}{{0.1}^2}=0.045\text{N}$
For charges at $B$ and $C$:
$F_2=\frac{9\times {10}^9\times 2.5\times {10}^{-7}\times 1\times {10}^{-7}}{{0.1}^2}=0.0225\text{N}$
We have taken the magnitudes and direction is shown in the figure. Since force is a vector quantity adding the two forces using parallelogram law of vector addition.
$F=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \alpha }$, here $\alpha$ is the angle between the two forces which is ${120}^{\circ }$.
Putting the value, we get,
$F=\sqrt{{0.045}^2+{0.0225}^2+2\times 0.045\times 0.0225\times \cos 120}$
$F=0.038\text{N}$