Question: Q. 5. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
U [Delhi I, II, III 2016]
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Solution:
Ans. Electric flux remains unaffected.
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[CBSE Marking Scheme, 2016]
Detailed Answer :
There will be no effect on electric flux, when the radius of Gaussian surface is increased because the charge enclosed by the gaussian surface remains the same.
The flux through Gaussian surface is given by $\phi=\frac{q_{e n c}}{\varepsilon_{0}}$. As the flux is independent of radius, so
there will be no effect on electric flux when radius of Gaussian surface increases.
[AI $Q$. 6. Two charges of magnitudes $-2 Q$ and $+Q$ are located at points $(a, 0)$ and $(4 a, 0)$ respectively. What is the electric flux due to these charges through a sphere of radius ’ $3 a^{\prime}$ with its centre at the origin?
U] [O.D., I, 2013]
Ans. Gauss’s theorem states that the electric flux through a closed surface enclosing a charge is equal to $\left(\frac{1}{\varepsilon_{0}}\right)$ times the magnitude of the charge enclosed.
The sphere encloses a charge of $-2 Q$, so
$$ \begin{equation*} \phi=\frac{2 Q}{\varepsilon_{0}} \tag{1} \end{equation*} $$
