Question

Charge q1 is distance r from a positive point charge q. charge q2=q1/3 is distance 2r from q. what is the ratio u1/u2 of their potential energies due to their interactions with q?

Answer 1

The ratio of the potential energies (U1 to U2) for charges q1 and q2, where q2 is q1/3 and at twice the distance from charge q, is calculated using the electric potential energy formula, resulting in a ratio of 6:1.

Step-by-step explanation:

The question involves determining the ratio of the potential energies (U1/U2) of two charges, q1 and q2, where q2 is a third the magnitude of q1, and at different distances from a positive point charge q. Using the formula for electric potential energy between two point charges, U = kq1q2/r, where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them, we can calculate the ratio.

For q1, the potential energy U1 is kqq1/r, and for q2, which is q1/3 and at a distance 2r, U2 is kq(q1/3)/(2r). Calculating these and taking the ratio U1/U2 gives us (kqq1/r) / (kq(q1/3)/(2r)) = 6.

Therefore, the ratio of their potential energies U1 to U2 is 6:1.

Answer 2

We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).

The potential energy of charge q is proportional to
∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹,

so if r₂ = 3r₁ and q₂ = q₁/4, then
U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁)
= 4•3 = 12.