Past exam of the physics 2 paper of the University of Oxford / 2015 / 9 / b

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- Solution b

Solution

Apply the negative gradient to the leading far-field electric potential. In spherical coordinates,

E_{r} = - \frac{\partial V}{\partial r} = \frac{2 α c o s θ}{r ^{3}}, E_{θ} = - \frac{1}{r} \frac{\partial V}{\partial θ} = \frac{α s i n θ}{r ^{3}}, E_{ϕ} = - \frac{1}{r s i n θ} \frac{\partial V}{\partial ϕ} = 0.

The electric field therefore has the coordinate-independent form

E = \frac{3 ( p \cdot r ^ ) r ^ - p}{4 π ϵ _{0} r ^{3}} .

For the electric dipole along

z

, use

V = α z / (x^{2} + y^{2} + z^{2})^{3/2}

. Direct partial differentiation gives the vector components

E_{x} = \frac{3 αx z}{r ^{5}}, E_{y} = \frac{3 α yz}{r ^{5}}, E_{z} = \frac{α ( 3 z ^{2} - r ^{2} )}{r ^{5}}, r^{2} = x^{2} + y^{2} + z^{2} .

On the positive axis the electric field points along the electric dipole moment; in the equatorial plane it points oppositely, providing a useful sign check.

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