JEE PHYSICS MCQ: CURRENT ELECTRICITY

1 Medium

1 of 1

Application +4 | -1

A parallel-plate capacitor of plate area

40 \mathrm{~cm}^2

and separation between the plates

0.10 \mathrm{~mm}

is connected to a battery of emf

2.0 \mathrm{~V}

through a

16 \Omega

resistor. Find the electric field in the capacitor

10 \mathrm{~ns}

after the connections are made.

View answer and explanation

Correct option: A

1.7 \times 10^{-4} \mathrm{~V/m}

Option A (

1.7 \times 10^{-4} \mathrm{~V/m}

) is correct because it is calculated using the charging equation for a capacitor and the relationship between electric field and voltage.

More Questions from CURRENT ELECTRICITY

Practice All Chapter MCQs →

\mathrm{\rho }

3\,\,{\mathrm{m}^3}

3\,\,\mathrm{\Omega }

4.7\,\, \times \,\,{10^6}\,\,\mathrm{\Omega }\,\, \pm \,\,5\,\,\%

22.5\,cm