In a region, the electric field is \(\vec E=500\,\hat i\ \mathrm{N\,C^{-1}}\) for \(x\gt0\) and \(\vec E=-500\,\hat i\ \mathrm{N\,C^{-1}}\) for \(x\lt0\). A closed cylinder of length 20 cm and diameter 20 cm has its axis along the x-axis and its centre at the origin. The net outward electric flux through the cylinder is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(31.4\ \mathrm{N\,m^{2}\,C^{-1}}\)
On the face at \( x=+10\ \mathrm{cm} \), the field points along \( +x \), the same as the outward normal. On the face at \( x=-10\ \mathrm{cm} \), the field points along \( -x \), which is again the outward normal. So each flat face has an outward flux \( E\pi r^2 \). The curved surface is parallel to the field and has no flux.
Given:
\( E=500\ \mathrm{N\,C^{-1}} \)
\( r=10\ \mathrm{cm}=0.1\ \mathrm{m} \)
\( \begin{aligned} \phi_{\text{face}} &= E\pi r^2 \\ &= 500\times3.14\times(0.1)^2 \\ &= 15.7\ \mathrm{N\,m^{2}\,C^{-1}} \end{aligned} \)
\( \begin{aligned} \phi &= 2\phi_{\text{face}} \\ &= 2\times15.7 \\ &= 31.4\ \mathrm{N\,m^{2}\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Example 1.11 – “(a) We can see from the figure that on the left face E and ΔS are parallel. …”
“On the right face, E and ΔS are parallel …”
“(b) For any point on the side of the cylinder E is perpendicular to ΔS and hence E.ΔS = 0. Therefore, the flux out of the side of the cylinder is zero.”