Two point charges q1=+2.45nC and q2=6.50nC are 0.100 m apart. Point A is midway

Question

Two point charges q1=+2.45nC 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.

Part A

Find the potential at point A.

Express your answer in volts to three significant figures.

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Part B

Find the potential at point B.

Express your answer in volts to three significant figures.

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Part C

Find the work done by the electric field on a charge of 2.70 nC that travels from point B to point A.

Express your answer in joules to two significant figures.

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Explanation / Answer

A) - Va= V(q1 on A)+V(q2 on A)= q1*9*10^9/(d1) +q2*9*10^9/(d2)..............if these particles in the air
= 2.45*10^-9 * 9*10^9 /0.05 + -6.5*10^-9 * 9*10^9 / 0.05 = -729 volt.
where : d1= the distance between q1 and A.
d2 = the distance between q2 and A.

B) - Vb= V(q1 on B)+V(q2 on A)= q1*9*10^9/(d3)+ q2*9*10^9/(d4)
= 2.45*10^-9 * 9*10^9/0.08 + -6.5*10^-9 * 9*10^9/0.06 = -699.38 volt

C)- work = qc*(Va-Vb)= 2.7*10^-9 *(-729-(-699.38))= -79.97*10^-9 J

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