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JEE Main 202227 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
27 June
Shift
Morning Shift
Subject
Physics
Chapter
Capacitance

JEE Main 2022 · 27 June · Morning Shift · Physics · Capacitance

A force of 10N acts on a charged particle placed between two plates...

A force of 10N10N 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 force of 10N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitorA force of 10N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor

Options

  1. A5N5N
  2. B10N10N
  3. C20N20N
  4. DZero

Correct answer

A. 5N5N

Step-by-step solution

A force of 10N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor

Electric field due to each plate, E1=E2=σ2ε0=Q2Aε0{E}_{1} = {E}_{2} = \dfrac{\sigma}{2{ε}_{0}} = \dfrac{Q}{2A{ε}_{0}}

Net electric field between the plates, Enet=E1+E2=QAε0{E}_{net} = {E}_{1} + {E}_{2} = \dfrac{Q}{A{ε}_{0}}

Force on charged particle between the plates, F1=qEnet=qQAε0=10N{F}_{1} = q{E}_{net} = \dfrac{qQ}{A{ε}_{0}} = 10N

A force of 10N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor

Now, in second case, the net electric field, E=σ2ε0=Q2Aε0=5NE = \dfrac{\sigma}{2{ε}_{0}} = \dfrac{Q}{2A{ε}_{0}} = 5N

Force on charged particle, F2=qE=qQ2Aε0=5N{F}_{2} = qE = \dfrac{qQ}{2A{ε}_{0}} = 5N

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