Answer:
The potential energy (U) of the system is 0.19778J
Explanation:
The potential energy (U) between two point charges (Q₁ and Q₂) is related to the force (F) of attraction/repulsion and the distance (r) between them as follows;
U = F x r ------------------(i)
Where, from Coulomb's law;
F = k x Q₁ x Q₂ ÷ r² ------- (ii)
k is the electric constant = 8.99 x N
<em>Substituting for F from equation(ii) in equation(i)</em>
=> U = (k x Q₁ x Q₂ ÷ r²) x r
=> U = k x Q₁ x Q₂ ÷ r ------------------(iii)
<em>Given;</em>
Q₁ = 2.0 μC = 2 x C
Q₂ = 5.5 μC = 5.5 x C
r = 0.50m
<em>Substitute the values of k</em><em>, r, Q₁ and Q₂ in equation (iii)</em>
=> U = 8.99 x x 2 x x 5.5 x ÷ 0.50
<em>Solve the equation;</em>
=> U = 8.99 x x 2 x x 5.5 x ÷ 0.50
=> U = 197.78 x
=> U = 0.19778J
<em>The potential energy (U) of the system is 0.19778J</em>