Two protons, each of charge #+1.60 * 10-19 C#, are #2.00 * 10-5 m# apart. What is the change in the electric potential energy of this pair of charges if they are brought #1.00 * 10-5 m# closer together?

Answer is A, why?

A) 1.15 10-23J
B) 3.20 10-19J
C) 3.20 10-16J
D) 1.60 10-19J
E) 1.60 * 10-14J

1 Answer
Jane
Mar 15, 2018

Initial electrostatic potential energy was #P.E=(9*10^9*(1.6*10^-19)^2)/(2*10^-5)J# (using the formula, #P.E=1/(4 pi epsilon) (q_1 q_2)/x#)

When they were brought closer by #1*10^-5m#,new distance between them became, #(2-1)*10^-5=1*10^-5m#

So,new potential energy is #P.E'=(9*10^9*(1.6*10^-19)^2)/(1*10^-5)J#

So,change in potential energy =#P.E'-P.E=(9*10^9*(1.6*10^-19)^2)/(10^-5)*(1-1/2)=11.52*10^-24=1.15*10^-23J#

Note here we get increase in potential energy during this process as bringing two positively charge results in repulsion,which causes the system to gain potential energy.

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