Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 6.52 x 10-12 m apart.

Question

Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 6.52 x 10^-12 m apart. What is EPE_final - EPE_initial, which is the change in the electric potential energy?

Answer 1

Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 6.52 x 10^-12 m apart. What is EPE_final - EPE_initial, which is the change in the electric potential energy?

using expression for Potential energy PE = K Q₁*Q₂/ r₁₂ This PE has to be negative as the charges are exerting force of attraction ( Initial PE =0 )

Change in PE in this case is -( 9*10⁹ *6*1.6* 10^-19 * 6* 10^-19 ) / 6.52 *10^-12 J

Answer 2

The initial PE is zero (because the charges are very far apart. )

The formula for the final PE is V = k q₁ q₂ / r₁₂

Noting the k = 9E9 , and the e is 1.6E-19, V_final= - 1.5E-15 J

Since one electron-volt is 1.6 E-19 J, V_final = - 7951 electron-volt