A charge Q = -110.0 nC is located 0.30 m frompoint A and 0.50 m from point B. (a

Question

A charge Q = -110.0 nC is located 0.30 m frompoint A and 0.50 m from point B. (a) What is the potential at A? V (b) What is the potential at B? V (c) If a point charge q is movedfrom A to B while Q isfixed in place, through what potential difference does it move? V Does its potential increase or decrease? (d) If q = -1.5 nC, what is the change in electric potentialenergy as it movesfrom A to B? J Does the potential energy increase or decrease? (e) How much work is done by the electric field due tocharge Q as q movesfrom A to B? J

Explanation / Answer

Q=-105*10^-9 C k=9*10^9 v=k*Q/r (a) What is the potential at A? va=9*10^9*(-105*10^-9)/0.3 = -3150 V (b) What is the potential at B? vb=9*10^9*(-105*10^-9)/0.5 = -1890 V c) The potential difference of b with respect to a vb-va=(-1890)-(-3150) = 1260 v Since positive work must be done to move the +q test charge from a to b: it's potential increases d) PE=(vb-va)*(-26*10^-9) 1260*(-26*10^-9) = -0.00003276 J it's potential energy decreases work done by the field e) Since q is positive, the field will not move q from A to Bwithout and external source of energy: work = deltav*q=1260*q J

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