A short electric dipole has a dipole moment of $16 imes 10^{-9} ext{ C m}$. The electric potential due to

Question

A short electric dipole has a dipole moment of $16 	imes 10^{-9} 	ext{ C m}$. The electric potential due to the dipole at a point at a distance of $0.6 	ext{ m}$ from the centre of the dipole situated on a line making an angle of $60^{\circ}$ with the dipole axis is: $(\frac{1}{4\pi\epsilon_0} = 9 	imes 10^9 	ext{ N m}^2/	ext{C}^2)$

Accepted Answer

The electric potential $V$ due to a short electric dipole at a distance $r$ from the centre and at an angle $	heta$ with the dipole axis is given by the formula:
$$V = \frac{1}{4\pi\epsilon_0} \frac{p \cos	heta}{r^2}$$ [NCERT Class 12, Physics Part I, Sec 2.4, Eq 2.15]

1.  **Given Data:**
    *   Dipole moment, $p = 16 	imes 10^{-9} 	ext{ C m}$
    *   Distance, $r = 0.6 	ext{ m}$
    *   Angle, $	heta = 60^{\circ}$
    *   Coulomb constant, $k = 9 	imes 10^9 	ext{ N m}^2	ext{C}^{-2}$

2.  **Calculation:**
    *   Substitute the values into the formula:
        $$V = \frac{(9 	imes 10^9) 	imes (16 	imes 10^{-9}) 	imes \cos(60^{\circ})}{(0.6)^2}$$
    *   Simplify the expression (note $10^9 	imes 10^{-9} = 1$ and $\cos 60^{\circ} = 0.5$):
        $$V = \frac{9 	imes 16 	imes 0.5}{0.36}$$
        $$V = \frac{144 	imes 0.5}{0.36} = \frac{72}{0.36}$$
        $$V = \frac{7200}{36} = 200 	ext{ V}$$

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