Two point charges q1=+2.60nC and q2=?6.30nC are 0.100 m apart. Point A is midway

Question

Two point charges q1=+2.60nC and q2=?6.30nC 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.

a.) Find the potential at point A. Express your answer in volts to three significant figures.

b.) Find the potential at point B. Express your answer in volts to three significant figures.

c.) Find the work done by the electric field on a charge of 2.50 nC that travels from point Bto point A. Express your answer in joules to two significant figures.

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
= 2.6*10^-9 * 9*10^9 /0.05 + -6.3*10^-9 * 9*10^9 / 0.05 = -666 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)
= 2.6*10^-9 * 9*10^9/0.08 + -6.3*10^-9 * 9*10^9/0.06 = -653volt
C)- work = qc*(Va-Vb)= 2.5*10^-9 *(-666)-(-653))= -32.5*10^-9 J

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