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Past exam of the physics 2 paper of the University of Oxford

2015

Apply the negative <gradient> to the leading far-field <electric potential>. In <spherical coordinates>,
$$
E_r=-\frac{\partial V}{\partial r}=\frac{2\alpha\cos\theta}{r^3},\qquad E_\theta=-\frac1r\frac{\partial V}{\partial\theta}=\frac{\alpha\sin\theta}{r^3},\qquad E_\phi=-\frac1{r\sin\theta}\frac{\partial V}{\partial\phi}=0.
$$
The <electric field> therefore has the coordinate-independent form
$$
\boxed{\mathbf E=\frac{3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r}-\mathbf p}{4\pi\epsilon_0r^3}.}
$$
For the <electric dipole> along $z$, use $V=\alpha z/(x^2+y^2+z^2)^{3/2}$. Direct <partial differentiation> gives the <vector components>
$$
\boxed{E_x=\frac{3\alpha xz}{r^5},\qquad E_y=\frac{3\alpha yz}{r^5},\qquad E_z=\frac{\alpha(3z^2-r^2)}{r^5},\quad r^2=x^2+y^2+z^2.}
$$
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.

Solution

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

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

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

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