Question

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)

Answer 1

Answer:

U=0.198J

Explanation:

The potential energy associated with two point charges is given as

U=(Kq1q2)/r

Where k=8.99*10⁹N.m²/C²

r=0.5m,

q₁=2μc,

q₂=5.5μc

If we substitute values into the equation we arrive at

U=(8.99*10⁹N.m²/C²* 2*10⁻⁶*5.5*10⁻⁶)/0.5

U=197.78*10⁻³

U=0.198J

From the calculation above, we can conclude that the potential energy between the two system of charges is 0.198J

Answer 2

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${e}$ x Q₁ x Q₂ ÷ r²  ------- (ii)

k${e}$ is the electric constant = 8.99 x $10^{9}$N$m^{2}/C^{2}$

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

=> U = (k${e}$ x Q₁ x Q₂ ÷ r²) x r

=> U = k${e}$ x Q₁ x Q₂ ÷ r   ------------------(iii)

Given;

Q₁ = 2.0 μC = 2 x $10^{-6}$ C

Q₂ = 5.5 μC = 5.5 x $10^{-6}$ C

r = 0.50m

Substitute the values of k${e}$, r, Q₁ and Q₂ in equation (iii)

=> U = 8.99 x $10^{9}$ x 2 x $10^{-6}$ x 5.5 x $10^{-6}$ ÷ 0.50

Solve the equation;

=> U = 8.99 x $10^{9}$ x 2 x $10^{-6}$ x 5.5 x $10^{-6}$ ÷ 0.50

=> U = 197.78 x $10^{-3}$

=> U = 0.19778J

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