Question

how much energy is needed to place four positive charges, each of magnitude +5.0 mc, at the vertices of a square of side 2.5 cm? choose one way of assembling the charges and calculate the potential at each empty vertex as this set of charges is assembled. be sure to clearly describe the order of assembly.

Answer 1

Assemble the charges in the order shown: 1  2 3 then 4:

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Initially, potential at corner #1 is zero. So energy to movecharge there is zero.

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Now there is a charge at corner #1. The potential at corner 2is:

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          V = k q1 / r = 8.99 x 10⁹ * 0.005 / 0.03525          note: diag of square is 3.525 cm

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           =    1.275 x 10⁹   Volts    (atcorner 2)

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Energy to bring in charge 2 is:

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       energy = qV =0.005 * 1.275 x 10⁹   =   6.376 x 10⁶ Joules

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Now potential at corner #3 is due to charges at 1 and 2:

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         V at 3 = k q1 / s + k q2 / s = 8.99 x10⁹ * 0.005 / 0.025 + 8.99 x 10⁹* 0.005 / 0.025 =

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           =    3.596 x 10⁹ Volts     (at corner 3)

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Energy to bring in charge to corner 3 is:

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       energy= qV = 0.005 *3.596x 10⁹ =   17.98 x 10⁶ Joules

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Potential now at corner 4 is due to charges at 1 2 and3:

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          3.596 + 1.275 =    4.871 x10⁹ Volts (at corner 4)

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   energy to bring in charge #4 = qV = 0.005 * 4.871 x 10⁹ =   24.355 x 10⁶ Joules

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total energy toassemble charges = potential energy stored =

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       = 24.355 + 17.98 + 6.376  =     48.71 x 10⁶ Joules    is total energy stored