A -3.0-μC point charge and a -9.0-μC point charge are initially extremely far apart. How much work does it take to bring the -3.0-μC charge to x = 3.0 mm, y = 0.00 mm, and the -9.0-μC charge to x = -3.0 mm, y = 0.00 mm? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2.)

It would take 40.5 Joules to bring the charges to the indicated points.

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The work required to bring the charges to the given points equals the electric potential energy of the system of two charges at these points. Since both charges are negative, the electrostatic force between them is repulsive, so the work has to be done against this force to bring the...

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The work required to bring the charges to the given points equals the electric potential energy of the system of two charges at these points. Since both charges are negative, the electrostatic force between them is repulsive, so the work has to be done against this force to bring the charges closer together.

Initially, the energy of the system of the two charges is zero, since they are very far apart. Once they are brought to the specified points, their energy will be

`U = ( k*q_1*q_2)/r_(12)` where `r_12` is the distance between the two charges.

Since the values of the `q_1` and `q_2` are both negative, their product is positive, so the energy will be positive. This is consistent with the positive work that has to be done to bring charges together.

The distance between the points can be found by examining their coordinates. Both points are located on the x-axis, so the distance between them is

`r_(12) = |x_2 - x_1| = |-3 - 3| = 6 mm` This can be converted to meters:

`r_12 = 6*10^(-3) m`

The charges, expressed in Coulombs, are `q_1 = -3*10^(-6) C` and `q_2 = -9*10^(-6) C`

The energy of the system of charges then will be

`U = (8.99*10^9*(-3*10^-6)*(-9*10^-6))/(6*10^(-3)) = 40.5 J`

This means that the work required to bring the charges to specified locations is 40.5 J.

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