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Tems11 [23]
3 years ago
8

The electrostatic force between two small charged particles is 5.2 x 10^-7 N at a distance of 2 meters. What will be the new for

ce if the distance is
increased to 4 meters apart?
Physics
1 answer:
Kamila [148]3 years ago
8 0

Answer:

F_2=1.3\times 10^{-7}\ N

Explanation:

Given that,

Initial force, F=5.2\times 10^{-7}\ N

distance, d = 2 m

New distance, d' = 4 m

We need to find the new force. The electrostatic force between two charges is given by :

F=k\dfrac{q_1q_2}{d^2}

So,

\dfrac{F_1}{F_2}=(\dfrac{d_2}{d_1})^2\\\\\dfrac{F_1}{F_2}=(\dfrac{4}{2})^2\\\\\dfrac{F_1}{F_2}=4\\\\F_2=\dfrac{F_1}{4}\\\\F_2=\dfrac{5.2\times 10^{-7}}{4}\\\\F_2=1.3\times 10^{-7}\ N

So, the new force is equal to 1.3\times 10^{-7}\ N.

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