Question

Charges $25Q,9Q$ and $Q$ are placed at points $ABC$ such that $AB=4m$, $BC=3m$ and angel between $AB$ and $BC$ is ${90}^o$. Then force on the charge $C$ is

Answer 1

Given that,

$AB=4$ m

$BC=3$ m

Now,

$AC=\sqrt{{\left(AB\right)}^2+{\left(BC\right)}^2}$

$AC=\sqrt{{\left(4\right)}^2+{\left(3\right)}^2}$

$AC=5m$

Now, the force on charge placed at $C$ by the charge on$A$

$F_{AC}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_A{q}_C}{r_{AC}^2}\right)N$

The force on charge placed at $C$ by the charge on $B$

$F_{BC}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_B{q}_C}{r_{BC}^2}\right)N$

So, both forces are perpendicular to each other

Now, the net force on charge at $C$

$F=\sqrt{{\left(F_{AC}\right)}^2+{\left(F_{BC}\right)}^2}$

$F=\frac{1}{4\pi {\epsilon }_0}\sqrt{\left(\frac{q_A{q}_C}{r_{AC}^2}\right)+\left(\frac{q_B{q}_C}{r_{BC}^2}\right)}$

$F=9\times {10}^9\sqrt{\left(\frac{25Q\times Q}{25}\right)+\left(\frac{9Q\times Q}{9}\right)}$

$F=9\sqrt{2}Q\times {10}^{9}N$

Hence, the force on charge $C$ is $9\sqrt{2}Q\times {10}^{9}N$