Initial Electric Potential Energy, Final Electric Potential Energy, and Work Done?

I would like some help in understanding how to solve a problem like this. There are four parts and the problem is as follows:

Consider a system of three charges. An object with charge q4 = +3.0 x 10^-9 C is moved to position C from infinity (not shown). q1 = q2 = q3 = +10.0 x 10^-8 C.

A) Determine the INITIAL electric potential of the system consisting of all four charges.

B) Determine the FINAL potential energy of the system consisting of all four charges.

C) Determine the work done on the system by moving the charge a4 from infinity to point C.

D) Determine the work done by the system by moving the charge q4 from infinity to point C, but for q1 = q3 = -10.0 x 10^-9 C and q2 = +10.0 x 10^-9 C.

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  • NCS
    Lv 7
    3 months ago
    Favorite Answer

    A) Determine the INITIAL electric potential of the system consisting of all four charges.

    In my opinion, this question doesn't make sense. Potential has to do with a specific location measured relative to another location. For instance, we usually assume that the potential at infinity is zero, and then consider some specific point relative to infinity.

    If the question is asking for the potential AT POINT C, then

    initial potential = algebraic Σ kQ/d = 8.99e9N·m²/C² * 3*10e-8C / 0.200m = 13 485 V

    HOWEVER, I suspect that your question MEANT to ask for the initial potential ENERGY, not the initial POTENTIAL.

    Initially, q4 has zero potential energy due to its distance. To find the potential energy of the three clustered charges, I usually consider (any) one of the charges initially situated there, and then I bring in the other specified charges one at a time. Potential energy = algebraic Σ kQq / d

    For instance, if you start with q1 in place and then you bring in q2, the new potential energy is

    PE = k*q1*q2 / (0.200m*√2)

    and if you then bring in q3, you have

    PE' = PE + k*q1*q3/0.400m + k*q2*q3/(0.200m*√2)

    or

    PE' = (k/0.200m)*[q1*q2/√2 + q1*q3/2 + q2*q3/√2]

    When I plug in your values for q1, q2 and q3 and use

    k = 8.99e9N·m²/C²

    I get

    PE' = 8.60e-4 J = 860 µJ = 0.860 mJ

    but check the math!

    B) Determine the FINAL potential energy of the system consisting of all four charges.

    So to PE' we need to add

    k*q4*[q1/0.200m + q2/0.200m + q3/0.200m] = -4.05e-5

    for q4 = -3.0e-9 C

    nd so

    PE" = 8.20e-4 J = 0.820 mJ = 820 µJ

    C) Determine the work done on the system by moving the charge a4 from infinity to point C.

    That would be -4.05e-5 J as shown in part B.

    D) Determine the work done by the system by moving the charge q4 from infinity to point C, but for q1 = q3 = -10.0 x 10^-9 C and q2 = +10.0 x 10^-9 C.

    Easiest is to consider that the potential at point C for the three initial charges is simply

    V = 8.99e9N·m²/C² * (-10.0 + 10.0 - 10.0)e-9C / 0.200m = 450 V

    and so the work done BY the system in bringing the fourth charge (positive) to position C is

    W = 450V * 10.0e-9C = 4.50e-6 J = 4.50 µJ

    Final note: I suspect that your initial values for q1 = q2 = q3 SHOULD have been +10.0 x 10^-9 C (not "10^-8 "). If that's the case, then you'll have to redo those calculations.

    Hope this helps!

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