A point charge q1 - -1-0 - C is held stationary at the origin- Asecond point charge q2 - - 2-0 - C moves from the point 1-0 m- 0- 0 to 2-0 m- 0- 0- How much work is done by the electric force on q2-

Work done = change in PEPE 1 = nbsp;1/4 πϵq_1 q_2/1 PE 2 = nbsp; nbsp;1/4 πϵq_1 q_2/2 work done = nbsp; nbsp;1/4 πϵq_1 q_2(1/1 1/2 ) = 9 m ...

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Potential Energy of a System of Multiple Point Charges

A point charg...

Question

A point charge $q_{1} = - 1.0 μ C$ is held stationary at the origin. Asecond point charge $q 2 = + 2.0 μ C$ moves from the point (1.0 m, 0, 0) to (2.0 m, 0, 0). How much work is done by the electric force on $q_{2}$ ?

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Solution

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q_{1} = + 2.40 μ C

q_{2} = - 4.30 μ

x = 0.150 m, y = 0

x = 0.250 m, y = 0.250 m .

q_{2}

- 356 \times 10^{- x} J

q_{1} = - 5.8 μ C

q_{2} = + 4.3 μ C

(0.26 m, 0, 0)

(0.38 m, 0, 0)

q_{2}

q_{1} = 1.0 μ C

q_{2} = - 2.0 μ C

10^{- 4}

189 \times 10^{- x} m s^{- 1}

q_{1}

q_{2}

- 6.4 \times 10^{- 8} J

4.2 \times 10^{- 8} J

q_{1} = 2 μ C

q_{2} = - 6 μ C

(0, 3 m)

(4 m, 0)

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