Two positive point charges are separated by a distance R . If the distance betwe

Question

Two positive point charges are separated by a distanceR. If the distance between the charges is reduced toR/2, what happens to the total electric potential energyof the system?

a. It is doubled.
b. It is reduced to one-half of itsoriginal value.
c. It increases by a factor of4.
d. It is reduced to one-fourth of itsoriginal value.
e. It remains the same. Two positive point charges are separated by a distanceR. If the distance between the charges is reduced toR/2, what happens to the total electric potential energyof the system? It is doubled. It is reduced to one-half of itsoriginal value. It increases by a factor of4. It is reduced to one-fourth of itsoriginal value. It remains the same.

Explanation / Answer

   Potentialenergy   U   =   (1/40)* q1 * q2 / r    Initially         U1   =   (1/40)* q1 * q2 / R    Finallay   U2   =   (1/40)* q1 * q2 /(R/2)   =   2 *(1/40) * q1 * q2 /R    Dividing   U2 /U1   =   2    i.e.         a. It isdoubled. It isdoubled.

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