Question

In the figure above, q1 = 3.00q and q2 = -q, where q = 0.01 µC, d1 = 7.13 cm, d2 = 10.70 cm, θ1 = 51.0 ° and θ2 = 63.0 °. How much work is done by the electric field as a charge Q = +4q is brought from infinity along the path to the point shown in the figure?

91 02 42 2,

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Answer #1

at infinity, Potential Vi = 0

at the given location,

Vf = k q1 / d1 + k q2 / d2

= ( 9 x 10^9)[ (3 x 0.01 x 10^-6)/(0.0713) + (-0.01 x 10^-6)/0.1070 ]

= 2946 Volt

Work done by electric field = - delta(PE)

= -q (VF - Vi )

= - 4 x 0.01 x 10^-6 x 2946

= - 117.8 x 10^-6 J

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