Question

$A B C$ is a right angled triangle. Calculate the magnitude of force on charge - $Q$ .

Accepted Answer

Force on charge $- Q$ due to $9 Q$

$F_{1} = \frac{1}{4 {πε}_{0}} \times \frac{|- Q| \times 9 Q}{{(3 a)}^{2}} = \frac{1}{4 {πε}_{0}} \times \frac{9 Q^{2}}{9 a^{2}}$

$\Rightarrow F_{1} = \frac{1}{4 π ε_{0}} \frac{Q^{2}}{a^{2}}$ towards $B A$

Force on charge $- Q$ due to $16 Q$

$F_{2} = \frac{1}{4 {πε}_{0}} \times \frac{|- Q| \times 16 Q}{{(B C)}^{2}}$

But, $B C = \sqrt{{(5 a)}^{2} - {(3 a)}^{2}} = 4 a$

$F_{2} = \frac{1}{4 {πε}_{0}} \times \frac{16 Q^{2}}{{(4 a)}^{2}} = \frac{1}{4 {πε}_{0}} \times \frac{16 Q^{2}}{16 a^{2}}$

$\Rightarrow F_{2} = \frac{1}{4 {πε}_{0}} \times \frac{Q^{2}}{a^{2}}$ towards $B C$

Net force on charge $- Q$ is the vector sum of forces $F_{1} a n d F_{2}$

$\Rightarrow F_{n e t} = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ}$

where, $θ = 90^{o}$ is the angle between forces $F_{1}$ and $F_{2}$

Also, $|F_{1}| = |F_{2}|$

$\Rightarrow F_{net} = \sqrt{2 F_{1}^{2} + 2 F_{1}^{2} \cos 90}$

$\Rightarrow F_{net} = \sqrt{2 F_{1}^{2}} = \sqrt{2} F_{1}$

$\Rightarrow F_{net} = \frac{\sqrt{2}}{4 {πε}_{0}} \times \frac{Q^{2}}{a^{2}}$

Embibe Experts Solutions for Chapter: Electrostatics, Exercise 3: BEGINNER'S BOX - 3

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Electrostatics, Exercise 3: BEGINNER'S BOX - 3

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