A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be :

Question

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be : (A) 5 N (B) 10 N (C) 20 N (D) Zero

Tardigrade Admin · Accepted Answer

F=q E=q((Q/A ∈0))=(q Q/A ∈0)=10 N Now, when one plate is removed.

E prime=(Q/2 A ϵ0) F=q E prime=(Q q/2 A ϵ0)=5 N

Q. A force of $10 N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be :

5 N

10 N

20 N

Zero

$F = qE = q (\frac{Q}{A \in _{0}}) = \frac{qQ}{A \in _{0}} = 10 N$
Now, when one plate is removed.

$E^{'} = \frac{Q}{2 A ϵ _{0}}$
$F = q E^{'} = \frac{Qq}{2 A ϵ _{0}} = 5 N$

Q. A force of 10N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be :

5 N

10 N

20 N

Zero

F=qE=q(A∈0​Q​)=A∈0​qQ​=10N Now, when one plate is removed. E′=2Aϵ0​Q​ F=qE′=2Aϵ0​Qq​=5N

Q. A force of $10 N$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be :

$F = qE = q (\frac{Q}{A \in _{0}}) = \frac{qQ}{A \in _{0}} = 10 N$
Now, when one plate is removed.

$E^{'} = \frac{Q}{2 A ϵ _{0}}$
$F = q E^{'} = \frac{Qq}{2 A ϵ _{0}} = 5 N$