Poll Results
No votes. Be the first one to vote.
Participate in Poll, Choose Your Answer.
Find the potential due the dipole when the angle subtended by the two charges at the point P is perpendicular.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
No votes. Be the first one to vote.
a
Explanation: The potential due the dipole is given by, V = m cos θ/(4πεr
2
). When the
angle becomes perpendicular (θ = 90). The potential becomes zero since cos 90 will
become zero.
The electric potential (V) due to a dipole at a point in space is given by the equation:
[ V = frac{k cdot p cdot cos(theta)}{r^2} ]
where:
– (k) is the Coulomb’s constant ((8.987 times 10^9 , text{Nm}^2/text{C}^2)),
– (p) is the magnitude of the dipole moment ((q cdot d), where (q) is the charge magnitude and (d) is the separation distance between the charges),
– (theta) is the angle between the dipole moment vector and the line joining the point in space to the midpoint of the dipole,
– (r) is the distance from the midpoint of the dipole to the point in space where the electric potential is being calculated.
When the angle ((theta)) subtended by the two charges at the point (P) is perpendicular, i.e., (theta = 90^circ), (cos(90^circ) = 0). Therefore, the potential (V) at point (P) due to the dipole, when (theta = 90^circ), is:
[ V = frac{k cdot p cdot cos(90^circ)}{r^2} = frac{k cdot p cdot 0}{r^