Get Answers to all your Questions

header-bg qa

Two plates are 2 cm  apart, a potential difference of 10 volt  is applied between them, the electric field between the plates is

  • Option 1)

    20\; \; N/C

  • Option 2)

    500\; \; N/C

  • Option 3)

    5\; \; N/C

  • Option 4)

    250\; \; N/C

 

Answers (1)

best_answer

As we learned

Electric Field Intensity -

\vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}

- wherein

 

 E=\frac{V}{d}=\frac{10}{2\times 10^{-2}}=500\; N/C

 


Option 1)

20\; \; N/C

Option 2)

500\; \; N/C

Option 3)

5\; \; N/C

Option 4)

250\; \; N/C

Posted by

Avinash

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE