Question

Two point charges q1=+2.55nC and q2=−6.80nC are 0.100 mm apart. Point AA is midway between them; point BB is 0.080 mm from q1 and 0.060 mm from q2. (See (Figure 1) .Take the electric potential to be zero at infinity.

part a)

Find the potential at point AA.

Express your answer in volts.

part b)

Find the potential at point B.

Express your answer in volts.

part c)

Find the work done by the electric field on a charge of 2.20 nC that travels from point B to point A.

Express your answer in joules to two significant figures.

Answer 1

a)

Distance between q1 and point AA =0.05mm

Potential at point AA due to q1=k x q1/0.00005

=8.9 x 10^9 x 2.55x10^-9 /0.00005

=4.539x10^5 V

Distance between q2 and point AA =0.05 mm

Potential at point AA due to q2 =-8.9x10^9x6.8x10^-9/0.00005

=-12.104x10^5 V

Potential at point AA=(4.539-12.104)x10^5

=-7.501x10^5 V

b)

Distance between q1 and point BB =0.08mm

Potential at point BB due to q1=k x q1/0.08 mm

=8.9 x 10^9 x 2.55x10^-9 /0.00008

=2.836x10^5 V

Distance between q2 and point BB=0.06 mm

Potential at point BB due to q2 =-8.9x10^9x6.8x10^-9/0.00006

=-10.087x10^5 V

Potential at point BB= (2.836-10.087) x10^5

=-7.251x10^5 V

c)

Work done by electricfield on charge 2.2 nC to shift charge from point AA to BB

=(-7.251 + 7.501) x10^5 x 2.2 x 10-9

=0.55x10-4 J