Question

Charges $5.0\times {10}^{-7}C,-2.5\times {10}^{-7}C$ and $1.0\times {10}^{-7}C$ are held fixed at the three corners $A,B,C$ of an equilateral triangle of side $10cm$ respectively. Find the electric force on the charge at $C$ due to the rest two.

Answer 1

Force on C due to A and B will be,

$F_{CA}=9\times {10}^9\times \frac{\left(5\times {10}^{-7}\right)\left(1\times {10}^{-7}\right)}{{\left(0.1\right)}^2}$

$=4.5\times {10}^{-2}N$

$F_{CB}=9\times {10}^9\times \frac{\left(-2.5\times {10}^{-7}\right)\left(1\times {10}^{-7}\right)}{{\left(0.1\right)}^2}$

$=-2.25\times {10}^{-2}N$

The angle between the two vectors will be ${120}^{\circ }$.

The resultant force due to the two forces will be,

$F=\sqrt{F_{CA}^2+F_{CB}^2+2F_{CA}{F}_{CB}\cos \theta }$

$=\sqrt{{\left(4.5\times {10}^{-2}\right)}^2+{\left(-2.25\times {10}^{-2}\right)}^2+2\left(4.5\times {10}^{-2}\right)\left(-2.25\times {10}^{-2}\right)\cos {120}^{\circ }}$

$=5.95\times {10}^{-2}N$