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mina [271]
3 years ago
5

2 equal charges, 27 micro Coulomb each, are separated by 5 cm. Find force between those.

Physics
1 answer:
padilas [110]3 years ago
3 0

Answer:

The force between charges is  F= 2.624*10^3N.

Explanation:

The Coulomb force F between the two charges q_1 and q_2 separated by distance d is given by the equation

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

where k is the coulombs constant, and has the value

k= 9*10^9N\cdot C\cdot m^2.

Now in our case

q_1=q_2=27\mu C =27*10^{-6}C

and

d= 5cm =0.05m,

therefore, the Coulomb force between the charges is

F =( 9*10^9N\cdot C\cdot m^2)*\dfrac{(27*10^{-6}C)(27*10^{-6}C)}{(0.05m)^2}

\boxed{ F= 2.624*10^3N}

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