The drawing shows a square, each side of which has a length of L = 0.250 m. Two different positive charges q1 and q2 are fixed at the corners of the square.

Question

The drawing shows a square, each side of which has a length of L = 0.250 m. Two different positive charges q₁ and q₂ are fixed at the corners of the square. Find the electric potential energy of a third charge q₃ = -5.00 x 10^-9 C placed at corner A and then at corner B.

q₁= 1.50 x 10^-9 C

q₂ = 4.00 x 10^-9 C

B-------------------A

I . . . . . . . . . . . . I

I . . . . . . . . . . . . I

I . . . . . . . . . . . . I

q₁-------------------q₂

EPE_A = ____ Units

EPE_B = _____ Units

Answer 1

As we know: EPE = q.V → EPE_A = q.V_A and EPE_B = q.V_B

Coulomb constant K = 1/(4πε_°) ≈ 9×10⁹ (N.m²)/C²

Electrostatic potential at point A due to q₁ and q₂

V_A = K q₁/(L.√2) + K(q₂/L) → V_A = (9×10⁹) [(1.5×10^-9)/0.354] + (9×10⁹) [(4.0×10^-9)/0.25] = 182.1 Volts

EPE_A = q₃.V_{A →} EPE_A = (-5.0×10^-9) (182.1) = -910.5 ×10^-9 J = -910.5 nJ

Electrostatic potential at point B due to q₁ and q₂

V_B = K (q₁/L) + Kq₂/(L√2) → V_B = (9×10⁹) [(1.5×10^-9)/0.25] + (9×10⁹) [(4.0×10^-9)/0.354] = 155.7 Volts

EPE_B = q₃.V_{B →} EPE_B = (-5.0×10^-9) (155.7) = -778.5 ×10^-9 J = -778.5 nJ

If q₁ = q₂ → V_A = V_B and also EPE_A = EPE_B

What is the meaning of EPE_A = -910.5 nJ?

When I am bringing q₃ from infinity to point A, I should use negative work.

Because the V_A is positive and q₃ will be accelerated toward point A.

But I should bring q₃ with a constant velocity.