Two protons, each of charge 1.60 · 10-19 C, are 2.00 · 10-5 m apart. What is the

Question

Two protons, each of charge 1.60 · 10-19 C, are 2.00 · 10-5 m apart. What is the change in potential energy if they are brought 1.00 · 10-5 m closer together? (ke = 8.99 · 109 N·m2/C2) .

Thanks, will rate Lifesaver & quickly!

I have answer but it is obtained when I do PE = ke(q1q2/r)   with using 2X10^-5 as r.  Since we want change in potential energy, wouldn't we calculate the PE using the new r and then subtract final PE from Initial PE?

Or use the diference in r's?  The answer given in the key is 1.15 x 10^-23J - maybe it is wrong?

I have found a time or two wrong answers in the key!  thanks....

Explanation / Answer

Charge of proton q = 1.6 x 10 -19 C Sepration of two protons r = 2 x 10 -5 m Potential when they are separated by a distance r is U = -Kqq / r Where k = Coulomb's constan t                = 8.99 x 10 9 N m 2 / C 2 Subsitute the values we get U = -1.15 x 10 -23 J Potential energy between two protons when they are separated by 1 x 10 -5 m is       U ' = -Kqq / r ' Where r ' = 1 x 10 -5 m Substitute values we get U ' = -2.301 x 10 -23 J Change in potential energy = U - U '                                          = 1.15 x 10 -23 J

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