A charge Q = -100.0 nC is located 0.30 m from point A and 0.50 m from point B .

Question

A charge Q = -100.0 nC is located 0.30 m from point A and 0.50 m from point B.

(a) What is the potential at A?
VA =   V
(b) What is the potential at B?
VB =   V

(c) If a point charge q is moved from A to B while Q is fixed in place, through what potential difference does it move?
?V =  V

Does its potential increase or decrease?
---Select--- decrease increase  

(d) If q = -19.5 nC, what is the change in electric potential energy as it moves from A to B?
?UE =   J

Does the potential energy increase or decrease?
---Select--- decrease increase  

(e) How much work is done by the electric field due to charge Q as q moves from A to B?
Wfield =  J

Explanation / Answer

Q=-100*10^-9 C
k=9*10^9
v=k*Q/r

(a) What is the potential at A?
va=9*10^9*(-100*10^-9)/0.3 = -3000 V

(b) What is the potential at B?
vb=9*10^9*(-100*10^-9)/0.5 = -1800 V

c)
The potential difference of b with respect to a
vb-va=(-1800)-(-3000) = 1200 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)*(-19.5*10^-9)
1200*(-19.5*10^-9) = -0.0000234 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 B without
and external source of energy:
work = deltav*q=1200*q J

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