Two point charges, Q 1 = +0.04 µC and Q 2 = -0.04 µC, are separated by adistance

Question

Two point charges, Q1 = +0.04 µC andQ2 = -0.04 µC, are separated by adistance d = 120 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 = 55 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     *
-100.69   OK

(b) What is the potential due to Q1 andQ2 at point B, located between the twocharges a distance a = 55 cm from the positive charge?

V(B) = V     *
100.69   OK

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

vB = m/s    

Explanation / Answer

you've got VA = -100.69 V and VB = 100.69V q = -1.6 x 10-19C, m = 9.1 x10-31kg energy at A is qVA energy at B is qVB + mv2/2 energy conservation: qVA = qVB + mv2/2 v = [2q(VA - VB)/m] = 8.42*106m/s

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