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:
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[tex]{e}[/tex] x Q₁ x Q₂ ÷ r² ------- (ii)
k[tex]{e}[/tex] is the electric constant = 8.99 x [tex]10^{9}[/tex]N[tex]m^{2}/C^{2}[/tex]
Substituting for F from equation(ii) in equation(i)
=> U = (k[tex]{e}[/tex] x Q₁ x Q₂ ÷ r²) x r
=> U = k[tex]{e}[/tex] x Q₁ x Q₂ ÷ r ------------------(iii)
Given;
Q₁ = 2.0 μC = 2 x [tex]10^{-6}[/tex] C
Q₂ = 5.5 μC = 5.5 x [tex]10^{-6}[/tex] C
r = 0.50m
Substitute the values of k[tex]{e}[/tex], r, Q₁ and Q₂ in equation (iii)
=> U = 8.99 x [tex]10^{9}[/tex] x 2 x [tex]10^{-6}[/tex] x 5.5 x [tex]10^{-6}[/tex] ÷ 0.50
Solve the equation;
=> U = 8.99 x [tex]10^{9}[/tex] x 2 x [tex]10^{-6}[/tex] x 5.5 x [tex]10^{-6}[/tex] ÷ 0.50
=> U = 197.78 x [tex]10^{-3}[/tex]
=> U = 0.19778J
The potential energy (U) of the system is 0.19778J
