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Question

Coulomb's Law
A $$7.50-nC$$ point charge is located $$1.80 \,m$$ from a $$4.20-nC$$ point charge.
(a) Find the magnitude of the electric force that one particle exerts on the other.
(b) Is the force attractive or repulsive?


Solution

(a) $$|F|=\dfrac{k_e|q_1||q_2|}{r^2}$$
$$F=\dfrac{k_e e^2}{r^2}=\dfrac{(8.99\times 10^{9} N.m^2/C^2)(7.50\times 10^{-9} C)(4.20\times 10^{-9}C)}{(1.80 \,m)^2}$$
$$=\boxed{8.74\times 10^{-8} \,N}$$
(b) The charges are like charges.$$\boxed{\text{The force is repulsive.}}$$

Physics
NCERT
Standard XII

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