# 2.7 and 2.11 Spherical Surface Charge

2.7 Find the electric field a distance $$z$$ from the centre of a spherical surface of radius $$R$$, which carries a uniform charge density of $$\sigma$$.

Let sphere be centered at the origin and its boundary defined by

$$x^2 + y^2 + z^2 = R^2$$

Let $$\vec{r}$$ be the position of a differential element of surface charge.

$$\vec{r} = x \, \hat{i} + y \, \hat{j} + \sqrt{ R^2 – x^2 – y^2 } \, \hat{k}$$

Let $$\vec{ H }$$ be the position of some point at which we wish to measure the electric field.

$$\vec{H} = z \, \hat{k}$$

$$\vec{R} = \vec{H} – \vec{r} = -x \, \hat{i} – y \, \hat{j} + \left( z – \sqrt{ R^2 – x^2 – y^2 } \, \right)\hat{k}$$

$$\left| \left| \vec{R} \right| \right| = R^2 + z^2 – 2 z \sqrt{ R^2 – x^2 – y^2 }$$

$$d\vec{E} = \frac{ 1 }{4 \pi \epsilon_0} \frac{ dq }{ \left( R^2 + z^2 – 2 z \sqrt{ R^2 – x^2 – y^2 } \right)^{3/2} } \left( – x \, \hat{i} – y \, \hat{j} + ( z – \sqrt{ R^2 – x^2 – y^2 }) \, \hat{k} \right)$$

Converting to spherical coordinates where $$\rho$$ is the radial distance from the origin.

$$\left| \left| \vec{R} \right| \right| = \rho^2 + z^2 – 2 z \sqrt{ R^2 – R^2 \sin^2 \phi \cos^2 \theta – R^2 \sin^2 \phi \sin^2 \theta }$$

$$\left| \left| \vec{R} \right| \right| = R^2 + z^2 – 2 z \sqrt{ R^2 – R^2 \sin^2 \phi } = R^2 + z^2 – 2 z R | \, \cos \phi \, |$$

$$d\vec{E} = \frac{ 1 }{4 \pi \epsilon_0} \frac{ dq}{ \left( R^2 + z^2 – 2 z R | \, \cos \phi \, | \right)^{3/2} } \left( – R \sin \phi \cos \theta \, \hat{i} – R \sin \phi \cos \theta \, \hat{j} + ( z – R | \, \cos \phi \, | ) \, \hat{k} \right)$$

$$d\vec{E} = \frac{ 1 }{4 \pi \epsilon_0} \frac{ \sigma R^2 \sin \phi \, d\theta \, d\phi}{ \left( R^2 + z^2 – 2 z R | \, \cos \phi \, | \right)^{3/2} } \left( – R \sin \phi \cos \theta \, \hat{i} – R \sin \phi \cos \theta \, \hat{j} + ( z – R | \, \cos \phi \, | ) \, \hat{k} \right)$$

$$\vec{E} = \frac{ \sigma R^2 }{4 \pi \epsilon_0} \int_0^{\pi} \int_0^{2 \pi} \frac{ \sin \phi \, d\theta \, d\phi}{ \left( R^2 + z^2 – 2 z R | \, \cos \phi \, | \right)^{3/2} } \left( – R \sin \phi \cos \theta \, \hat{i} – R \sin \phi \cos \theta \, \hat{j} + ( z – R | \, \cos \phi \, | ) \, \hat{k} \right)$$

$$\vec{E} = \frac{ 2 \pi \sigma R^2 }{4 \pi \epsilon_0} \int_0^{\pi} \frac{ \left( z – R | \, \cos \phi \, | \right) \sin \phi \, d\phi}{ \left( R^2 + z^2 – 2 z R | \, \cos \phi \, | \right)^{3/2} } \, \hat{k}$$

Using Mathematica

$$\vec{E} = \frac{ 2 \pi \sigma R^2 }{4 \pi \epsilon_0} \left( \frac{ z – R }{z^2 | \, z – R \, |} + \frac{ z + R }{ z^2 | \, z + R \, |} \right) \, \hat{k}$$

For $$z > R$$

$$\vec{E} = \frac{ 2 \pi \sigma R^2 }{4 \pi \epsilon_0} \left( \frac{ 1 }{z^2} + \frac{ 1 }{ z^2 } \right) \, \hat{k} = \frac{ \sigma R^2}{ z^2 \epsilon_0} \, \hat{k} = \frac{ Q_{total} }{4 z^2 \pi \epsilon_0} \, \hat{k}$$

For $$z < R$$ $$\vec{E} = \frac{ 2 \pi \sigma R^2 }{4 \pi \epsilon_0} \left( - \frac{ 1 }{z^2} + \frac{ 1 }{ z^2 } \right) \, \hat{k} = 0 \, \hat{k}$$

>p>

Due to the symmetry using the integral form of Gauss’ law is far more efficient (question 2.11).

$$\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$$

for $$\rho < R$$ $$\left| \left| \vec{E} \right| \right| 4 \pi \rho^2 = \frac{Q_{enc}}{\epsilon_0}$$ $$\left| \left| \vec{E} \right| \right| = \frac{Q_{enc}}{4 \pi \epsilon_0 \rho^2}$$ Since $$Q_{enc} = 0$$ then $$\left| \left| \vec{E} \right| \right| = 0$$

for $$\rho > R$$

$$\left| \left| \vec{E} \right| \right| 4 \pi \rho^2 = \frac{Q_{enc}}{\epsilon_0}$$

$$\left| \left| \vec{E} \right| \right| = \frac{Q_{enc}}{4 \pi \epsilon_0 \rho^2}$$

$$Q_{enc} = \sigma ( 4 \pi R^2 )$$

$$\left| \left| \vec{E} \right| \right| = \frac{\sigma (4 \pi R^2)}{4 \pi \epsilon_0 \rho^2} = \frac{\sigma R^2}{\epsilon_0 \rho^2}$$