Problem 18.04 A point charge q 1 = +2.40 ? C is held stationary at the origin. A

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Problem 18.04 A point charge q1= +2.40?C is held stationary at the origin. A second point charge q2= -4.30?C moves from the point x = 0.170m , y = 0 m , to the point x = 0.235m , y = 0.235m . Part A How much work is done by the electric force on q2? Problem 18.04 A point charge q1= +2.40?C is held stationary at the origin. A second point charge q2= -4.30?C moves from the point x = 0.170m , y = 0 m , to the point x = 0.235m , y = 0.235m . Problem 18.04 A point charge q1= +2.40?C is held stationary at the origin. A second point charge q2= -4.30?C moves from the point x = 0.170m , y = 0 m , to the point x = 0.235m , y = 0.235m . A point charge q1= +2.40?C is held stationary at the origin. A second point charge q2= -4.30?C moves from the point x = 0.170m , y = 0 m , to the point x = 0.235m , y = 0.235m . A point charge q1= +2.40?C is held stationary at the origin. A second point charge q2= -4.30?C moves from the point x = 0.170m , y = 0 m , to the point x = 0.235m , y = 0.235m . Part A How much work is done by the electric force on q2? Part A How much work is done by the electric force on q2? Part A How much work is done by the electric force on q2? Part A How much work is done by the electric force on q2? Part A How much work is done by the electric force on q2? Part A How much work is done by the electric force on q2? Problem 18.04 A point charge q1= +2.40?C is held stationary at the origin. A second point charge q2= -4.30?C moves from the point x = 0.170m , y = 0 m , to the point x = 0.235m , y = 0.235m . Part A How much work is done by the electric force on q2?

Explanation / Answer

Workdone = change in potential * Charge

Change in potential = 9*10^9*10^-6*(2.4/0.17 - 2.4 / (0.235*sqrt(2))) = 62065.179 V

Workdone = 62065.179 * 4.3*10^-6 = 0.26688 J

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