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Electric Charges and Fields — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields3 Marks
Question
Write Gauss's law and explain.
✓

Answer

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→ Let us consider the total flux through a sphere of radius $r$, which encloses a point charge $q$ at its centre. Divide the sphere into small area elements, as shown in fig.
→ The flux through an area element $\Delta S$ is
$\Delta \phi=\overrightarrow{ E } \cdot \Delta \overrightarrow{ S }=\frac{q}{4 \pi \varepsilon_0 r^2} \hat{r} \cdot \overrightarrow{\Delta S }$
→ The unit vector $\hat{r}$ is along the radius vector from the centre to the area element.
→ Now, since the normal to a sphere at every point is along the radius vector at that point, the area element $\Delta S$ and $\hat{r}$ have the same direction.
Therefore,
$\Delta \phi=\frac{q}{4 \pi \varepsilon_0 r^2} \Delta S$
→ The total flux through the sphere is obtained by adding up flux through all the different area elements:
→ Now S, the total area of the sphere, equals $4 \pi r^2$. Thus,
$\phi=\frac{q}{4 \pi \varepsilon_0 r^2} \times 4 \pi r^2=\frac{q}{\varepsilon_0}$
→ Equation (1) is a simple illustration of a general result of electrostatics called Gauss's law.
→ "The total electric flux associated with a closed surface is equal to the ratio of total enclosed charge and $\varepsilon_0$."
$\therefore \phi=\int \overrightarrow{ E } \cdot d \overrightarrow{ S }=\frac{\sum q}{\varepsilon_0}$
$\phi=\sum_{\text {avl } \Delta S } \frac{q}{4 \pi \varepsilon_0 r^2} \Delta S$
→ Since each area element of the sphere is at the same distance $r$ from the charge,
$\phi=\frac{q}{4 \pi \varepsilon_0 r^2} \sum_{\text {all } \Delta S } \Delta S =\frac{q}{4 \pi \varepsilon_0 r^2} S$

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