The <span>roles of products, reactants, and limiting reactant are significant in order for a chemical reaction to take place. The reactants are the components that are interacted in such a way that it would eventually produce a product, while on the other hand, a limiting reactant is used to regulate the products intended to produce.</span>
Answer:

Explanation:
As we know that the charge per unit length of the long cylinder is given as

here we know that the electric field between two cylinders is given by

now we know that electric potential and electric field is related to each other as





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
