Two point charges of values +2.0 and +5.5 μC, respectively, are separated by 0.50 m. What is the potential energy of this 2-charge system? (ke = 8.99 × 109 N⋅m2/C2)

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

Substituting for F from equation(ii) in equation(i)

=> U = (k x Q₁ x Q₂ ÷ r²) x r

=> U = k x Q₁ x Q₂ ÷ r   ------------------(iii)

Given;

Q₁ = 2.0 μC = 2 x C

Q₂ = 5.5 μC = 5.5 x C

r = 0.50m

Substitute the values of k, r, Q₁ and Q₂ in equation (iii)

=> U = 8.99 x x 2 x x 5.5 x ÷ 0.50

Solve the equation;

=> U = 8.99 x x 2 x x 5.5 x ÷ 0.50

=> U = 197.78 x

=> U = 0.19778J

The potential energy (U) of the system is 0.19778J

it was a fall down when it happens

explanation:

answer:

(b)

explanation: