Question

The electric potential due to a dipole at point P making an angle \(120^{\circ }\) with the axis of a dipole is
(\(\varepsilon _{0}\) = permittivity of free space, \(p\) = electric dipole moment , \(r\) = distance of point P from the center of dipole)
(\(\cos 60^{\circ }=1/2\))

• A. \(\frac{1}{8\pi \varepsilon _{0}}\frac{p}{r^{2}}\)
• B. \(\frac{-1}{8\pi \varepsilon _{0}}\frac{p}{r^{2}}\)
• C. \(\frac{-1}{8\pi \varepsilon _{0}}\frac{3p}{r^{2}}\)
• D. \(\frac{1}{8\pi \varepsilon _{0}}\frac{3p}{r^{2}}\)

Hint:

Use the dipole potential V equals p cos theta divided by 4 pi epsilon zero r squared and put theta equal to 120 degrees.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The potential of a short dipole at distance \(r\) and angle \(\theta\) from its axis is \(V=\frac{1}{4\pi\varepsilon_{0}}\frac{p\cos\theta}{r^{2}}\).

Step 2: Calculation
For \(\theta=120^{\circ}\), \(\cos120^{\circ}=-\cos60^{\circ}=-\frac{1}{2}\).

Step 3: Result
\(V=\frac{1}{4\pi\varepsilon_{0}}\frac{p}{r^{2}}\left(-\frac{1}{2}\right)=-\frac{1}{8\pi\varepsilon_{0}}\frac{p}{r^{2}}\). It is negative because P lies on the side of the negative charge.

Final Answer:
The potential is minus p over 8 pi epsilon zero r squared. \[ \boxed{\text{(B) }-\dfrac{1}{8\pi\varepsilon_{0}}\dfrac{p}{r^{2}}} \]