Force acting upon a charged particle kept between the plates of charged capacitor is $F$ . If one of the plates of the capacitor is removed, force acting on the same particle will become :

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{F}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{F}{2}$
Electric field between the oppositely charged plates of a capacitor is twice of that due to one plate. Hence, when one plate is removed, the electric force reduces to half of its earlier value.

Suggest Corrections

Similar questions

Force acting upon a charged particle kept between the plates of a charged condenser is F. If one plate of the condenser is removed, then the force acting on the same particle will become:

Force acting upon a charged particle kept between the plates of a charged condenser is F.If one of the plates of the condenser is removed,force acting on the same particle will become

Q. Force acting on a test charge between the plates of a parallel capacitor is

F

. If one of the plates is removed, then the force on the same test charge will be:

Q. Two charged particles exert a force of magnitude

F

on one another. If the distance between them is doubled and the charge of one of the particles becomes four times, Find out the new force acting between them?

Q. In a parallel-plate capacitor of plate area A, plate separation d and charge Q, the force of attraction between the plates is F.

Join BYJU'S Learning Program

Force acting upon a charged particle kept between the plates of charged capacitor is F. If one of the plates of the capacitor is removed, force acting on the same particle will become :

The correct option is B F2Electric field between the oppositely charged plates of a capacitor is twice of that due to one plate. Hence, when one plate is removed, the electric force reduces to half of its earlier value.

Force acting upon a charged particle kept between the plates of charged capacitor is $F$ . If one of the plates of the capacitor is removed, force acting on the same particle will become :

The correct option is B $\frac{F}{2}$
Electric field between the oppositely charged plates of a capacitor is twice of that due to one plate. Hence, when one plate is removed, the electric force reduces to half of its earlier value.