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    

(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    

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

vB = m/s    

Explanation / Answer

(a) Electric potential at A = VA = K[Q2/a +Q1/(d-a)] (b) Electric potential at B = VB = K[Q1/a +Q2/(d-a)] (c) speed of the elctron would be such that its kineticenergy at B = loss of potential energy as compared to that atA.     potential energy at A = UA=e*VA= ........     potential energy at B =UB= e*VB= ........     so, E = UB - UA= 1/2*m*v*v = .........      so, v = (2E/m) =.........

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