" I Two point chasge "q_(1)=10times10^(-8)c,8_(2)=-2times10^(-8)c

Two point charge q_(1) = 10 xx 10^(-8)C and q_(2) = -2 xx 10^(-8)C are separated by a distance of 6 cm in air. (ii) The electrostatic potential energy of the system is

Two point charge q_(1) = 10 xx 10^(-8)C and q_(2) = -2 xx 10^(-8)C are separated by a distance of 6 cm in air. (i) The distance from q_(1) of the point of zero electric potential is

Two points charges q_1 = 10 xx 10^-8 c and q_2 = -2 xx 10^-8 C are separated by a distance of 60cm in air. Also calulate the electrostatic potential energy of the system.

Two points charges q_1 = 10 xx 10^-8 c and q_2 = -2 xx 10^-8 C are separated by a distance of 60cm in air. Find at what distance form the charge, would the electric potential be zero.

What is the electrostatic potential energy of the charge configuration shown in figure. Take q_(1)=+1.0xx10^(-8)C,q_(2)=-2.0xx10^(-8)C,q_(3)=+3.0xx10^(-8)C,q_(4)=+2.0xx10^(-8)C and a=1.0 metre.

Four point charges q_(1),q_(2),q_(3)" and " q_(4) are placed at the corners of the squares of side a,as shown in figure.Calculate the potential at the centre of the square. (Given, q_(1)=1xx10^(-8)C,q_(2)= -2xx10^(-8)C,q_(3)=3xx10^(-8)C,q_(4)=2xx10^(-8)C" and " a=1m )

23.54times10= (c) 8.352times100=

A 15pF capacitor is connected to a 60V battery.How much electrostatic energy is stored? (A) 2.70times10^(-8)J (B) 1.77times10^(-8)J (C) 2.35times10^(-8)J (D) 1.35times10^(-8)J