Two point charges q1 = 2.30 nC and q2 = -6.60 nC are 0.100 m apart. Point A is m

Question

Two point charges q1 = 2.30 nC and q2 = -6.60 nC 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 (the figure (Figure 1) ). Take the electric potential to be zero at infinity. Part A Find the potential at point A. VA = V SubmitMy AnswersGive Up Part B Find the potential at point B. VB = V SubmitMy AnswersGive Up Part C Find the work done by the electric field on a charge of 3.00 nC that travels from point B to point A. Express your answer using two significant figures. W =

Explanation / Answer

Here ,

as the potential due to a charge is given as

V = k * q/r

a) for the potential at A

potential at A = 9 *10^9 * 10^-9 *(2.30/0.050 - 6.6/.050)

potential at A = -774 V

b) at the potential at B

potential at B = 9 *10^9 * 10^-9 * (2.30/.080 - 6.6/.060)

potential at B = -731 V

c)

for the work done

work done in moving the charge = charge * change in potential

work done in moving the charge = 3 *10^-9 * (-731 - (-774))

work done in moving the charge = 1.29 *10^-7 J

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