Two point charges, Q 1 = +0.01 µC and Q 2 = -0.01 µC, are separated by adistance

Question

Two point charges, Q1 = +0.01 µC andQ2 = -0.01 µC, are separated by adistance d = 80 cm. An electron (q = -1.6 x10-19C, m = 9.1 x 10-31kg) isreleased from rest from a point A located between the twocharges a distance a = 35 cm from the negative charge, andmoves in a straight line toward the positive charge.

(a) What is the potential at point A due to the chargesQ1 and Q2? (Take thepotential at infinity to be zero.)

V(A) = V    

(b) What is the potential due to Q1 andQ2 at point B, located between the twocharges a distance a = 35 cm from the positive charge?

V(B) = V    

(c) What is the speed of the electron when it passes pointB?

vB = m/s

Explanation / Answer

a), Let the potential at pointA be V1 V1=kQ2/a+kQ1/(d-a) V1=-57,14(V) b) Let the potential at point B be V2. V2=kQ2/(d-a)+kQ1/a V2=57,14(V) c) energy is conservative so we have e*V1=e*V2+mv^2/2 e*(V1-V2)=mv^2/2 1,6e-19*57,14*2=9,1e-31*v^2/2 so v=6,34e6(m/s)

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