Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
111 views
in CBSE by (15 points)
Derive an expression of the potential due to an electric dipole at a point lies on its axial position where r>a provided that the potential due to an infinity is 6 V

Please log in or register to answer this question.

1 Answer

0 votes
by (54.6k points)

\(V_{net} = V_1 + V_2\)

\(= -\frac1{4\pi \epsilon_0} \frac q{r + a} + \frac 1{4\pi \epsilon_0} \frac q{r - a}\)

\(= \frac q{4\pi \epsilon_0}\left(\frac 1{r-a} - \frac 1{r + a}\right)\)

\(= \frac 1{4\pi \epsilon_0} q\left(\frac{r + a- r + a}{r^2 -a^2}\right)\)

\(= \frac 1{4\pi \epsilon_0} \frac{q.2a}{r^2 - a^2}\)

\(\because p = q.2 a\)

\(= \frac 1{4\pi \epsilon_0} \frac{p}{r^2 - a^2}\)

For r > a

\(v = \frac 1{4 \pi \epsilon_0} \frac p{r^2}\)

Related questions

0 votes
1 answer
asked Mar 3 in CBSE by OK BRO (20 points)
0 votes
2 answers
asked Oct 7, 2021 in CBSE by Afra Raffi (20 points)
0 votes
0 answers
asked Sep 10, 2021 in CBSE by adarsh8967 (15 points)
0 votes
0 answers
asked Sep 7, 2020 in CBSE by Angeleena Manju (15 points)
0 votes
1 answer
asked Feb 27, 2020 in CBSE by Vishal000069 (15 points)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...