Need help: Two particles with charges +6.00 e and-5.00 e are initially very far

Question

Need help:
Two particles with charges +6.00e and-5.00e are initially very far apart (effectively aninfinite distance apart). They are then fixed at positions that are6.67 x 10-12m apart. What is EPEfinal -EPEinitial, which is the change in the electricpotential energy?
Two particles with charges +6.00e and-5.00e are initially very far apart (effectively aninfinite distance apart). They are then fixed at positions that are6.67 x 10-12m apart. What is EPEfinal -EPEinitial, which is the change in the electricpotential energy?

Explanation / Answer

   EPE   =   (1/40)* q1 * q2 / r    given   q1   =   +6.00 e   =   +6.00 * 1.6 *10-19   =   96.0 *10-19   C    q2   =   -5.00 e   =   - 5.00 * 1.6 *10-19   =   - 80.0 *10-19   C    rinitial   =   ,   rfinal   =   6.67* 10-12   m    EPEfinal   -   EPEinitial   =   9.0* 109 * [{ 96.0 * 10-19 * ( -80.0 *10-19) / } -   {96.0 *10-19 * ( -80.0 * 10-19) / 6.67 *10-12}]    Change inenergy      =   + 1.036 * 10-13   J

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