Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Two point charges A and B of magnitude \mathrm{+8 \times 10^{-6} \mathrm{C}} and  \mathrm{-8 \times 10^{-6} C} respectively are placed at a distance d apart. The electric field at the middle point O between the charges is \mathrm{6.4 \times 10^{4} \mathrm{NC}^{-1}}. The distance 'd' between the point charges A and B is :

Option: 1

\mathrm{2.0 \mathrm{~m}}


Option: 2

\mathrm{3.0 \mathrm{~m}}


Option: 3

\mathrm{1.0 \mathrm{~m}}


Option: 4

\mathrm{4.0 \mathrm{~m}}


Answers (1)

Electric field at 'P'

\mathrm{E=\frac{Kq}{\left ( \frac{d}{2} \right )^{2}}}\times 2=8\frac{Kq}{d^{2}}

So,

\mathrm{E=\frac{8\times 9\times 10^{9}\times 8\times 10^{-6}}{d^{2}}}=6.4\times 10^{4}

         \mathrm{d^{2}=9}

         \mathrm{d=3m}

Hence option 2 is correct.

Posted by

Ramraj Saini

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

Bihar Board extends BSEB Super 50 free coaching application last date to March 26
anam-khan

JEE Main 2025: Most important chapters for physics, chemistry, mathematics to score good marks
anam-khan

JEE Main 2025 session 2 dates clash with ISC, PSEB 12th exams; no impact on major science subjects

Latest Question