Question

# Coulomb's LawA $$7.50-nC$$ 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.$$\boxed{\text{The force is repulsive.}}$$PhysicsNCERTStandard XII

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