Question

A point charge with a charge q1 = 2.20 ?C is held stationary at the origin....

A point charge with a charge q1 = 2.20 ?C is held stationary at the origin. A second point charge with a charge q2 = -4.90 ?C moves from the point x= 0.110 m , y= 0 to the point x= 0.250 m , y= 0.250 m . ow much work is done by the electric force on q2?

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Answer #1

Solution:

Answer : -0.54 N/m

Charge q1 = 2.2 x10-6 C

q2 = - 4.9 x 10-6 C

Initial position = r1 = 0.11 m

Inital electroststic potential energy = U1 =- k q1 q2 /r1 = -(9 x109)( 2.2 x10-6) (-4.9 x10-6) / (0.11)

= 0.882 N /m

Final distance =r2 = (x2-x1)2 +(y2-y1)2 = (0.25 -0.11)2 + (0.25 - 0)2 = 0.2865 m

Final electrostatic potential energy = U2 = - k q1 q2 / r2 = -(9 x109)( 2.2 x10-6) (-4.9 x10-6) / (0.2865)

= 0.3386 N/m .

Work done on q2 = Change in electrostatic potential energy = U2 - U1 = 0.3386 - 0.882

= - 0.54 N/m .

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