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

What school do you go to?

Step-by-step explanation:

Answer 2

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

Step-by-step 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=(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 (k(3.5)(-6))/(x^2)=(k(5)(-6))/((40-x)^2)`

Canceling similar terms

`\Rightarrow \left [ (40-x)/(x)\right ]^2=(10)/(7)\\\\\Rightarrow (40-x)/(x)=\sqrt{(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.