Q. 615.0( 4 Votes )

Two charged particles, having equal charges of 2.0 × 10–5 C each, are brought from infinity to within a separation of 10 cm. Find the increase in the electric potential energy during the process.

Answer :


Given:
Charge of two particles: q1=q2= 2.0 × 10–5 C
Separation between two charges: r = 10 cm = 0.1 m
Formula used:
Electric potential energy is given as:
Where,k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is the separation between two charges.
When two charges were at infinity, the separation between them was infinite
Thus,

Now, when the separation between them was 10 cm;
Where Uf is the final electric potential energy.
Substituting the values we get,

Now, increase in electric potential energy: Δ U


Hence, Electric potential energy increased by 36 J during the process.


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