Answer:
(a) the force is 8.876 N
(b) the magnitude of each charge is 4.085 μC
Explanation:
Part (a)
Given;
coulomb's constant, K = 8.99 x 10⁹ N.m²/C²
distance between two charges, r = 10 cm = 0.1 m
force between the two charges, F = 15 N
when the distance between the charges changes to 13 cm (0.13 m)
force between the two charges, F = ?
Apply Coulomb's law;

Part (b)
the magnitude of each charge, if they have equal magnitude

where;
F is the force between the charges
K is Coulomb's constant
Q is the charge
r is the distance between the charges

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