KCET2020PhysicsElectrostatics
The electric field lines on the left have twice the separation on those on the right as shown in figure. If the magnitude of the field at A is 40 Vm ⁻¹ , what is the force on 20 C charge kept at B ?
Options
- A4 10⁻⁴ ~N
- B8 10⁻⁴ ~N
- C16 10⁻⁴ ~N
- D1 10⁻⁴ ~N
Correct answer
A. 4 10⁻⁴ ~N
Step-by-step solution
According to given figure, Electric field at point A, E_ A =40 Vm ⁻¹ Since, electric field lines on the left have twice the separation on those on the right (at point B ), hence electric field at point B aligned &E_ B = E_ A 2 = 40 2 &E_ B =20 Vm ⁻¹ aligned Force on charge q kept at B is gathered F=q E_ B Given, q=20 C =20 10⁻⁶ C F=20 10⁻⁶ 20=4 10⁻⁴ ~N gathered