Electric Field and Voltage due to an infinitely long thick slab inside and outside the slab

 

Electric Field and Voltage due to an infinitely long thick slab inside and outside the slab

Electric Field outside an Infinitely Long Slab

Assumption (Symmetry)

We consider the origin at the center of the slab.

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Electric field due to an elemental slab

$$dE = \frac{\sigma}{2\varepsilon_0}$$

For a thin slab of thickness $dx$,

$$dq = \rho \, A \, dx$$
$$\sigma = \frac{dq}{A} = \frac{\rho \, A \, dx}{A} = \rho \, dx$$

So,

$$dE = \frac{\rho \, dx}{2\varepsilon_0}$$

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Net Electric Field Inside the Slab

Integrating from $x = -\frac{L}{2}$ to $x = +\frac{L}{2}$:

$$E = \int dE = \int_{-L/2}^{L/2} \frac{\rho}{2\varepsilon_0} \, dx$$
$$E = \frac{\rho}{2\varepsilon_0} \int_{-L/2}^{L/2} dx$$
$$E = \frac{\rho}{2\varepsilon_0} \cdot L$$

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Final Result

$$E_{\text{outside slab}} = \frac{\rho L}{2\varepsilon_0}$$

(constant)

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Relation with Potential

$$E = -\frac{dV}{dx}$$ $$-\frac{dV}{dx} = \frac{\rho L}{2\varepsilon_0}$$ $$dV = -\frac{\rho L}{2\varepsilon_0} \, dx$$

Integrating:

$$V = -\frac{\rho L}{2\varepsilon_0} x + C$$

Note

• $C$ is a constant determined using boundary/reference conditions.

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Electric Field Inside a Thick Infinite Slab (At a Distance $x$)

Concept

• Consider an infinitely long uniformly charged slab.

• Let the origin $x = 0$ be at the center.

• At a distance $x$, only the charge enclosed within $-x$ to $+x$ contributes to the electric field.

• Charges outside this region cancel due to symmetry.

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Electric Field Inside the Slab

From symmetry and previous result:

$$E = \frac{\rho}{2\varepsilon_0} \cdot (2x)$$
$$E = \frac{\rho x}{\varepsilon_0}$$

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Final Result

$$E_{\text{inside slab}} = \frac{\rho x}{\varepsilon_0}$$

• Electric field is not constant

• It increases linearly with distance $x$
  

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Relation with Potential

$$E = -\frac{dV}{dx}$$ $$-\frac{dV}{dx} = \frac{\rho x}{\varepsilon_0}$$ $$dV = -\frac{\rho x}{\varepsilon_0} \, dx$$

Integrating:

$$V = -\frac{\rho}{\varepsilon_0} \int x \, dx$$
$$V = -\frac{\rho}{2\varepsilon_0} x^2 + C$$

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Final Potential Expression

$$V = -\frac{\rho x^2}{2\varepsilon_0} + C$$

• This represents an inverted parabola

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Key Observations

• $E \propto x$ → linear variation

• $V \propto -x^2$ → parabolic variation

• At center $(x = 0)$:

  $$E = 0, \quad V = \text{maximum}$$
  

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