# 2.17 Infinite Slab

2.17 An infinite plane slab, of thickness $$2d$$, carries a uniform volume charge density $$\rho$$. Find the electric field as a function of $$y$$, where $$y=0$$ at the centre. Plot $$\vec{E}$$ vs $$y$$,

Let $$\vec{r}$$ be the position of an arbitrary differential volume of charge within the slab.

$$\vec{r} = x \, \hat{i} + y \, \hat{j} + z \, \hat{k}$$

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

$$\vec{H} = y_0 \, \hat{k}$$

$$\vec{R} = \vec{H} – \vec{r} = – x \, \hat{i} + (y_0 – y) \, \hat{j} + z \, \hat{k}$$

$$d\vec{E} = \frac{1}{4 \pi \epsilon_0 } \frac{ \rho \, dx \, dz \, dy }{ ( x^2 + (y_0 – y)^2 + z^2)^{3/2} } ( – x \, \hat{i} + (y_0 – y) \, \hat{j} + z \, \hat{k} )$$

$$\vec{E} = \frac{ \rho }{ 4 \pi \epsilon_0 } \int_{-d}^d \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{ dx \, dz \, dy }{ ( x^2 + (y_0 – y)^2 + z^2)^{3/2} } ( – x \, \hat{i} + (y_0 – y) \, \hat{j} + z \, \hat{k} )$$

Since the $$\hat{i}$$ and $$\hat{k}$$ components involve convergent integrals whose integrands are odd functions about symmetric limits of integration they will equal zero, leaving only.

$$\vec{E} = \frac{ \rho }{ 4 \pi \epsilon_0 } \int_{-d}^d \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{ (y_0 – y) \, dx \, dz \, dy }{ ( x^2 + (y_0 – y)^2 + z^2)^{3/2} } \, \hat{j}$$

$$x = \sqrt{ z^2 + (y_0 – y)^2} \tan \theta \hspace{30mm} dx = \sqrt{ z^2 + (y_0 – y)^2} \sec^2 \theta \, d\theta$$

$$\vec{E} = \frac{ \rho }{ 4 \pi \epsilon_0 } \int_{-d}^d \int_{-\infty}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{ (y_0 – y) \, \sqrt{ z^2 + (y_0 – y)^2} \sec^2 \theta \, d\theta}{ ( (y_0 – y)^2 + z^2)^{3/2} ( \sec^2 \theta )^{3/2} } \, dz \, dy \, \hat{j}$$

$$\vec{E} = \frac{ \rho }{ 4 \pi \epsilon_0 } \int_{-d}^d \int_{-\infty}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{ (y_0 – y) \cos \theta \, d\theta}{ ( (y_0 – y)^2 + z^2) } \, dz \, dy \, \hat{i} = \frac{ 2 \rho }{ 4 \pi \epsilon_0 } \int_{-d}^d \int_{-\infty}^{\infty} \frac{ (y_0 – y)}{ ( (y_0 – y)^2 + z^2) } \, dz \, dy \, \hat{j}$$

$$z = (y_0 – y) \tan \phi \hspace{30mm} dz = (y_0 – y)\sec^2 \phi \, d\phi$$

$$\vec{E} =\frac{ 2 \rho }{ 4 \pi \epsilon_0 } \int_{-d}^d \int_{-\pi/2}^{\pi/2} \frac{ (y_0 – y)^2 \sec^2 \phi \, d\phi}{ (y_0 – y)^2 \sec^2 \phi } \, dy \, \hat{j} = \frac{ 2 \pi \rho }{ 4 \pi \epsilon_0 } \int_{-d}^d dy \, \hat{j} = \frac{ \rho d }{ \epsilon_0 } \, \hat{j}$$

The above answer assumes that $$y_0$$ is outside the slab since the limits of integration for $$dy$$ are $$(-d, d)$$. Which implies that the field is constant everywhere!