AP EAMCET201823 Apr 2018Morning ShiftPhysicsElectrostaticsActual
Two equally charged metal spheres A and B repel each other with a force of 4 10⁻⁵ ~N . Another identical uncharged sphere C is touched to A and then placed at the mid- point of the line joining the spheres A and B . The net electric force on the sphere C is
Options
- A4 10⁻⁵ ~N from C to A
- B4 10⁻⁵ ~N from C to B
- C8 10⁻⁵ ~N from C to A
- D8 10⁻⁵ ~N from C to B
Correct answer
A. 4 10⁻⁵ ~N from C to A
Step-by-step solution
Force, F= k q^2 d^2 =4 10⁻⁵ ~N Now, A is touched by C , then; Charge on C=q / 2 Charge on A=q / 2 So, force on C= F _A+ F _B aligned & = k q / 2 q / 2 (d / 2)^2 r _ A C + k q q / 2 (d / 2)^2 r _ B C & .= k q^2 / 4 d^2 / 4 - k q^2 / 2 d^2 / 4 as r _ A C =- r _ B C ] & = k q^2 d^2 (1-2)=- k q^2 d^2 =-4 10⁻⁵ ~N aligned So, force is of same magnitude but in opposite direction, i.e from C to A as suggested by minus sign.