) from a +5.0-nC A +3.0-nC c charge q fixed at the origin. Naturally, the Coulom
ID: 1573156 • Letter: #
Question
) from a +5.0-nC A +3.0-nC c charge q fixed at the origin. Naturally, the Coulomb force accelerates Q o, eventually reaching 15 cm (r2). harge Q is initially at rest a distance of 10 cm ( away from Q = +3.0 nC q = +5.0 nC r1 = 10 cm r2 = 15 cm The charge Q is repelled by q, thus having work done on it and losing potential energy. What is the work done by the electric field between r, and r2? What is the change in the potential energy of the two-charge system from r, to r2? How much kinetic energy does Q have at r2?Explanation / Answer
F = k Q q / r^2
and W = integral of F.dr from r1 to r2
W = (k Q q )(-1/r2 + 1/r1)
W = (9 x 10^9 x 5 x 10^-9 x 3 x 10^-9) (-1/0.15 + 1/0.10)
W = 4.5 x 10^-7 J .......Ans
change in PE = - Work done = - 4.5 x 10^-7 J ..........Ans
from energy conservation,
this loss of Pe will convert into KE of Q.
so Ke at r2 = 4.5 x 10^-7 J ......Ans
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.