Question

A short electric dipole has a dipole moment of \(16 \times 10^{-9} C\, m\). The electric potential due to the dipole at a point at a distance of \(0.6 \,m\) from the centre of the dipole, situated on a line making an angle of \(60\degree\)with the dipole axis is : \((\frac {1}{4\pi \epsilon_o}= 9 \times 10^9 N\,m^2/C^2)\)

• A. $50\,V$
• B. $200\,V$
• C. $400\,V$
• D. zero

Answer 1

The Correct Option is B

To solve the problem of finding the electric potential due to a dipole at a given point, we will use the formula for the electric potential \( V \) due to an electric dipole. For a point at a distance \( r \) from the center of a dipole with dipole moment \( p \) and at an angle \( \theta \) with the dipole axis, the potential is given by:

\(V = \frac{1}{4\pi \epsilon_0} \cdot \frac{p \cdot \cos \theta}{r^2}\)

Let's substitute the given values into the formula:

1. The dipole moment \( p = 16 \times 10^{-9} \, \text{C m} \).
2. The distance \( r = 0.6 \, \text{m} \).
3. The angle \( \theta = 60^\degree \).
4. The constant \( \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

Substituting these values, we get:

\(V = \left(9 \times 10^9 \right) \cdot \frac{16 \times 10^{-9} \cdot \cos(60^\degree)}{(0.6)^2}\)

Now, calculate \( \cos(60^\degree) = \frac{1}{2} \).

Substituting \( \cos(60^\degree) \) gives:

\(V = 9 \times 10^9 \cdot \frac{16 \times 10^{-9} \cdot \frac{1}{2}}{0.36}\)

\(V = 9 \times 10^9 \cdot \frac{8 \times 10^{-9}}{0.36}\)

Calculating further,

\(V = 9 \times \frac{8}{0.36} = 200 \, \text{V}\)

Hence, the electric potential at the given point is \(200 \, \text{V}\).

Thus, the correct option is \(200 \, \text{V}\).