The electric potential due to a dipole at point P making an angle $120^{}^{\circ}$ with the axis of a dipole is
(\varepsilon_0 $= permitivity of free space,$ p $= electric dipole moment ,$ r $= distance of point P from the center of dipole)<br>(\cos 60^{}^{\circ} = 1/2$ )

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

= permitivity of free space,

p

= electric dipole moment ,

r

= distance of point P from the center of dipole)<br>(\cos 60^{}^{\circ} = 1/2

)

Quantrex Academy · Accepted Answer

Correct answer: B. (-1)/(8π _0)(p)/(r^2)

The electric potential V due to a dipole at a distance r and angle θ from the dipole axis is given by V = (1)/(4π _0) (p θ)/(r^2) Given θ = 120^ ^ , we have 120^ ^ = - 60^ ^ = -(1)/(2) Substituting this value into the formula, we get V = (1)/(4π _0) (p (-(1)/(2)))/(r^2) V = (-1)/(8π _0) (p)/(r^2) Answer: (-1)/(8π _0)(p)/(r^2)

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