Question

A charge of +3.5 nC and a charge of +5.0 nC are separated by 40 cm. Find the equilibrium position for a -6.0 nC charge.

Answer 1

Answer:

What school do you go to?

Explanation:

Answer 2

Answer: $18.22\ cm$ from $3.5\ nC$ charge.

Explanation:

Given

The magnitude of the first charge is  $Q_1=3.5\ nC$

The magnitude of the second charge is $Q_2=5\ nC$

$-6\ nC$ charge must be placed in between the two charges to establish equilibrium

The electrostatic force is given by

$F=\dfrac{kq_1q_2}{r^2}$

Equilibrium will be established when force by both the charges balance out each other. Suppose $-6\ nC$ is placed at a distance of $x$ cm from $3.5\ nC$ . So, we can write

$\Rightarrow \dfrac{k(3.5)(-6)}{x^2}=\dfrac{k(5)(-6)}{(40-x)^2}$

Canceling similar terms

$\Rightarrow \left [ \dfrac{40-x}{x}\right ]^2=\dfrac{10}{7}\\\\\Rightarrow \dfrac{40-x}{x}=\sqrt{\dfrac{10}{7}}=1.195\\\\\Rightarrow 40-x=1.195x\\\Rightarrow 40=2.195x\\\Rightarrow x=18.22\ cm$

Thus, the equilibrium position is $18.22\ cm$ from $3.5\ nC$ charge.