Question: Q. 3. State Gauss’s law in electrostatics.
R [Delhi I, II, III 2016]
Show Answer
Solution:
Ans. Gauss’s law in electrostatics : “The surface integral of electrostatic field $\vec{E}$ produced by any source over any closed surface $S$ enclosing a volume $V$ in vacuum, i.e., total electric flux over the closed surface $S$ in vacuum, is $\frac{1}{\varepsilon_{0}}$ times the total charge $(\mathrm{Q})$ contained inside $\mathrm{S}$, i.e.,
$$ \begin{equation*} \phi_{\mathrm{E}}=\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}}=\frac{\mathrm{Q}{\text {enclosed }}}{\varepsilon{0}} \tag{1} \end{equation*} $$
[CBSE Marking Scheme, 2016]
Commonly Made Error
- Most of the candidates are unable in stating gauss law correctly. Key words like net charges, closed surface etc. are missed by students. Some candidates write magnetic flux instead of electric flux.
