Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Deduce an expression for the potential energy of a system of two point charges q1 and q2 located at position r1 and r2 respectively in an external field (E) 

Answers (1)

Work done to bring q1 from infinity to r1 = q1v(r1)

Work done to bring q2 from infinity to r2 = q2v(r2)

Also, work done on q2 to move it against the field due to \mathrm{q}_{1}=\frac{k q_{1} q_{2}}{r_{12}}
Where,
\mathrm{k}=\frac{1}{4 \pi \epsilon_{o}}
r_{12}= distance between q1 and q2
. . potential energy of the system =\mathrm{q}_{1} \mathrm{v}\left(\mathrm{r}_{1}\right)+\mathrm{q}_{2} \mathrm{v}\left(\mathrm{r}_{2}\right)+\frac{k q_{1} q_{2}}{r_{12}}

Posted by

Satyajeet Kumar

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

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question