Find the equipotential surface

There are two factors that contribute to the electric potential of this configuration: the external electric field and the dipole. We set up a coordinate system as follows: Place the dipole at the origin, with its moment points right along the $x$ -axis ( $\vec{p}=p\hat{x}$ ). The external electric field also points along the positive $x$ -axis.

The potential of the electric field is $V(x) = -E_{0}x$ where we choose $x=0$ to be the point of zero potential. Remember that electric field points to the direction of decreasing potential, hence the minus sign.

The potential of the dipole is given by the formula, but since $\vec{p}$ only has $x$ -components, only the $x$ -component of the position will matter for the dot product. Since the configuration we will be studying is spherically symmetric, $r$ is basically constant. Thus the potential for the dipole is given by $V(x) = \frac{kpx}{r^{3}}$

The total potential at any point is $V_{total}(x)= x(\frac{kp}{r^{3}}-E_{0})$

In order for a sphere to be equipotential, potential cannot depend on $x$ . The only way this is possible is if the expression in the parentheses above is zero and $V=0$ everywhere. Thus,

$\frac{kp}{r^{3}}-E_{0} = 0$

$\frac{kp}{r^{3}}=E_{0}$

$r^{3} = \frac{kp}{E_{0}} = \frac{(9*10^{9})(1.2*10^{-8})}{1}$

$r^{3} = 108$

$r=4.762\textrm{m}$

The electric potential created by the dipole is given by: $\Phi(\vec{r})=\frac{1}{4\pi\varepsilon_0}\frac{\vec{p} \cdot \vec{r}}{r^3}$ In the other hand, the electric potential created by the constant electric field is given by: $\Phi(\vec{r})=\vec{E_0} \cdot \vec{r}$ The total electric potential is then the sum of both: $\Phi(\vec{r})=\frac{1}{4\pi\varepsilon_0}\frac{\vec{p} \cdot \vec{r}}{r^3}+\vec{E_0} \cdot \vec{r}$ For simplicity, without loss of generality, we can make $\vec{p} = p\; \hat{i}$ and $\vec{E_0} = E_0\; \hat{i}$ using $\vec{r} = x \; \hat{i} + y \; \hat{j} + z \; \hat{k}$ as the position vector. In this case turns out that: $\vec{p} \cdot \vec{r} = p y$ And $\vec{E_0} \cdot \vec{r} = E_0 y$ So that the potential field can be written as: $\Phi(\vec{r})=\frac{1}{4\pi\varepsilon_0}\frac{p y}{(x^2+y^2+z^2)^{3/2}}+E_0 y$ Assuming that for a certain $\Phi = t$ the shape of this equipotential is a sphere we can write: $t=\frac{1}{4\pi\varepsilon_0}\frac{p y}{(x^2+y^2+z^2)^{3/2}}+E_0 y$ $\frac{1}{4\pi\varepsilon_0}\frac{p y}{(x^2+y^2+z^2)^{3/2}}=t-E_0 y$ $(x^2+y^2+z^2)^{3/2}=\frac{1}{4\pi\varepsilon_0}\frac{p y}{t-E_0 y}$ If the shape is a sphere then we must have $x^2+y^2+z^2=R²$ and then $(x^2+y^2+z^2)^{3/2}$ must be a constant. Differentiating the both sides with respect y yields: $\frac{d}{dy}(x^2+y^2+z^2)^{3/2}=\frac{d}{dy}(\frac{1}{4\pi\varepsilon_0}\frac{p y}{t-E_0 y})$ $\frac{d}{dy}\frac{1}{4\pi\varepsilon_0}\frac{p y}{t-E_0 y} =0$ $\frac{p(t-E_0y)+E_0py}{(t-E_0y)^2} =0$ $t-E_0y=-E_0y$ $t=0$ Now we know the spherical equipotential is given for $\Phi = 0$ and then we can solve the equation to find the radius: $\frac{1}{4\pi\varepsilon_0}\frac{p y}{(x^2+y^2+z^2)^{3/2}}=-E_0 y$ $(x^2+y^2+z^2)^{3/2}=\frac{1}{4\pi\varepsilon_0}\frac{p}{-E_0}$ $(x^2+y^2+z^2)^{3}=(\frac{1}{4\pi\varepsilon_0}\frac{p}{E_0})^2$ $(x^2+y^2+z^2)^{1/2}=(\frac{1}{4\pi\varepsilon_0}\frac{p}{E_0})^{2/3}$ $R = (\frac{1}{4\pi\varepsilon_0}\frac{p}{E_0})^{2/3}$ Finally using the values given in the problem: $R = (9\times 10^9\; m/F \frac{1.2\times 10^{-8} \;m \cdot C}{1\; V/m})^{2/3}$ $R = 4.762 \;m$

Clearly , the surface should cut field lines perpendicularly . Hence , the component of field parallel to surface should be 0. Let $\theta$ be the angle between $\vec{r}$ and $\vec{p}$ . We know that component of electric field due to dipole parallel to surface or perpendicular to $\vec{r}$ = $k\frac{psin\theta}{R^3}$ . Component of the external electric field along the same axis( perpendicular to $\vec{r}$ ) = $E_{0}sin\theta$ . net field along this axis = 0 , $\Rightarrow k\frac{psin\theta}{R^3} = E_{0}sin\theta$ $\Rightarrow R^3 = k\frac{p}{E_{0}} = 108$ , hence $R = \fbox{4.762}$

We will consider the center of the dipole to be the origin, and let $\vec{r} =(x,y,z)$ be the position vector of a general point. Assuming the dipole to be in the direction of the x-axis, the electric dipole moment will be represented by the vector $\vec{p} = (p,0,0)$

The electric potential created by the dipole at the general point is $\phi_{dipole} = k\frac{\vec{r}\vec{p}}{r^3} = \frac{kxp}{\left( \sqrt{x^2+y^2+z^2}\right)^3}$

Furthermore there is the electric potential created by the electric field. Let's assume the electric field is in the decreasing x-direction and potential created by the field is $0$ at the origin, we would have $\phi_{field} = -E_0x = -x$

Thus the total electric potential is $\phi=\frac{kxp}{\left( \sqrt{x^2+y^2+z^2}\right)^3}-x$

Now we consider the equipotential surface where $\phi=0$ : $\frac{kxp}{\left( \sqrt{x^2+y^2+z^2}\right)^3}-x=0$

We divide by both sides by $x$ and square both sides to get $\frac{(kp)^2}{(x^2+y^2+z^2)^3}-1=0$ , or $(x^2+y^2+z^2)^3=(kp)^2$ Taking the cube root of both sides, we get $x^2+y^2+z^2=(kp)^{2/3}$

This is the equation of a sphere centered at the origin and radius $r^2 = (kp)^{2/3}$ , or $r=(kp)^{1/3}=(9\times 10^9 \times 1.2 \times 10^{-8})^{1/3}=4.762 \mathrm{ m}$

Consider a point P on the sphere shown:

Alternate text

Now For equipotential surface, Potential due to Dipole = Potential due to External Field

(Let R make an angle $\theta$ with dipole)

Potential due to Dipole = $\frac { \overrightarrow { p } \cdot \quad \overrightarrow { R } }{ 4\pi { \varepsilon }_{ 0 }{ R }^{ 3 } } =\frac { p\quad cos\theta }{ 4\pi { \varepsilon }_{ 0 }{ R }^{ 2 } }$

Potential due to External Field = $\oint { \overrightarrow { E } \cdot \overrightarrow { dR } } ={ E }_{ 0 }\quad R\quad cos\theta$

Now Equate both, to get - ${ E }_{ 0 }\quad R\quad cos\theta \quad =\quad \frac { p\quad cos\theta }{ 4\pi { \varepsilon }_{ 0 }{ R }^{ 2 } }$

thus, $R=\quad \sqrt [ 3 ]{ \frac { 1.2\quad \times \quad 9\quad \times \quad 10 }{ 1 } } =\quad 4.762\quad m\quad$ [using calculator]