Two point charges q 1=+2.40nC and q 2=?6.50nC are 0.100 m apart. Point A is midw

Question

Two point charges q1=+2.40nC and q2=?6.50nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q1 and 0.060 m from q2. (See (Figure 1) .) Take the electric potential to be zero at infinity.

Find the potential at point A.

Express your answer in volts to three significant figures.

Find the potential at point B.

Express your answer in volts to three significant figures.

Find the work done by the electric field on a charge of 2.50 nC that travels from point B to point A.

Express your answer in joules to two significant figures.

Explanation / Answer

electric potential due to a charge Q at a distance r is V = kQ/r

where k = 8.9875*10^9 Nm/C^2

(a) At point A

V = k [(q1/r)+(q2/r)] = (8.9875*10^9)*[(2.4*10^-9)+(-6.5*10^-9)]/0.05 = - 736.975 v

(b) At point B

V = k [(q1/r1)+(q2/r2)] = (8.9875*10^9)*[((2.4*10^-9)/0.08)+((-6.5*10^-9)/0.06)] = 704.020 v

(c) workdone = force * displacement = F*d = E*q*d = (k*q1*q*d)/r^2

=> workdone = [(8.9875*10^9)*(2.4*10^-9)*(2.5*10^-9)*0.08]/((0.08)^2) = 674.06*10^-9 J

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