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lilavasa [31]
3 years ago
10

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.

Physics
2 answers:
Alisiya [41]3 years ago
7 0

Answer:

What school do you go to?

Explanation:

Zinaida [17]3 years ago
5 0

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.

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