Question

A charge of 10 e.s.u. is placed at a distance of 2 cm from a charge of 40 e.s.u. and 4 cm
rom another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs)​

Answer 1

A charge of 10 e.s.u. is placed at a distance of 2 cm from a charge of 40 e.s.u. and 4 cm rom another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs)

solution : potential energy of the charge 10 e.s.u = potential energy between 10 e.s.u and 40 e.s.u + potential energy between 10 e.s.u and 20 e.s.u

= k(10e.s.u)(40e.s.u)/(2cm) + k(10e.s.u)(20e.s.u)/(4cm)

value of k in mks = 9 × 10^9 Nm²/C²

but value of k in cgs = 1 dyne cm²/e.s.u²

so, potential energy of the charge 10e.s.u = 10 × 40/2 + 10 × 20/4

= 200 + 50

= 250 erg

hence, potential energy of the charge 10 e.s.u = 250 erg

Answer 2

Answer:

Explanation:potential energy of the charge 10 e.s.u = potential energy between 10 e.s.u and 40 e.s.u + potential energy between 10 e.s.u and -20 e.s.u

= k(10e.s.u)(40e.s.u)/(2cm) + k(10e.s.u)(-20e.s.u)/(4cm)

value of k in mks = 9 × 10^9 Nm²/C²

but value of k in cgs = 1 dyne cm²/e.s.u²

so, potential energy of the charge 10e.s.u = 10 × 40/2 - 10 × 20/4

= 200 - 50

= 150 erg

Hope you got it