Two charged particles, each carrying a charge of 2.0 10⁻⁵ C , are moved from infinity to a separation of 10 cm . Determine the increase in the electric potential energy during this process.

Step-by-step solution

The initial electric potential energy of the system when the charges are at infinity is U_i = 0 . The final electric potential energy when the charges are separated by a distance r is given by: U_f = 1 4 ₀ q₁ q₂ r Given q₁ = q₂ = 2.0 10⁻⁵ C and r = 10 cm = 0.1 m . Substituting the values: U_f = (9 10^9) (2.0 10⁻⁵) (2.0 10⁻⁵) 0.1 U_f = 9 10^9 4.0 10⁻¹⁰ 0.1 U_f = 3.6 0.1 = 36 J The increase in electric potential energy is U = U_f - U_i = 36 J .