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GUJCET · Physics · Electric Charges and Fields
When \(10 \mu C\) charge is enclosed by a closed surface, the flux passing through the surface is \(\phi\). Now another \(10 \mu C\) charge is placed inside the r closed surface, then the flux passing through the surface is _________.
- A \(4 \phi\)
- B \(\phi\)
- C \(2 \phi\)
- D zero
Answer & Solution
Correct Answer
(C) \(2 \phi\)
Step-by-step Solution
Detailed explanation
(C)
\(\begin{array}{l}q=10 \mu C \\ \phi=\frac{q}{\varepsilon_0}\end{array}\)
In \(2^{\text {nd }}\) condition total charge enclosed by a closed surface is
\(
\begin{array}{l}
Q=10+10=20 \mu C \\
Q=2 q
\end{array}
\)
Total flux passing through the surface is
\(\begin{aligned} \phi^{\prime} & =\frac{ Q }{\varepsilon_0}=\frac{2 q}{\varepsilon_0} \\ \therefore \quad \phi^{\prime} & =2 \phi \text { (From equation }(1))\end{aligned}\)
\(\begin{array}{l}q=10 \mu C \\ \phi=\frac{q}{\varepsilon_0}\end{array}\)
In \(2^{\text {nd }}\) condition total charge enclosed by a closed surface is
\(
\begin{array}{l}
Q=10+10=20 \mu C \\
Q=2 q
\end{array}
\)
Total flux passing through the surface is
\(\begin{aligned} \phi^{\prime} & =\frac{ Q }{\varepsilon_0}=\frac{2 q}{\varepsilon_0} \\ \therefore \quad \phi^{\prime} & =2 \phi \text { (From equation }(1))\end{aligned}\)
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