Question

A charge of 10 esu is placed at a distance of 2 cm
from a charge of 40 esu and 4 cm from another
charge of - 20 esu. The potential energy of the
charge 10 esu is :- (in ergs)
1) 87.5
2)112.5
3)150
4)zero​

Answer 1

Answer:

please mark and follow

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

= 250 erg

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