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Two point charges each experience a 1-N electrostatic force when they are 2 cm apart. If they are moved to a new separation of 8 cm, what is the magnitude of the electric force on each of them?

2 N

1/8 N

1/16 N

1/4 N

1/2 N

User Luqi
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5.3k points

1 Answer

3 votes

Electrostatic force between two points in space is defined as,


F_e=(Q_1Q_2)/(4\pi r\epsilon_r\epsilon_0)

The r is the distance between them.

So if,


1N=(Q_1Q_2)/(4\pi 2cm\epsilon_r\epsilon_0)\Rightarrow 2\cdot1N=(Q_1Q_2)/(4\pi\cdot10^(-2)m\cdot\epsilon_r\epsilon_0)

Than,


\boxed{(1)/(4)N}=(Q_1Q_2)/(4\pi\cdot 8cm\cdot\epsilon_r\epsilon_0)\Rightarrow8(1)/(4)N\Leftrightarrow 2N=(Q_1Q_2)/(4\pi\cdot10^(-2)m\cdot\epsilon_r\epsilon_0)

Hope this helps.

r3t40

User Frarees
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5.2k points