Question

Option 1: What is Gauss's law? The figure shows a cylinder placed in an electric field \(\vec{E}=100\,\hat{i}\ \text{N/C}\). \(\vec{E}\) is positive when \(x>0\) and negative when \(x<0\). Length of the cylinder is \(30\ \text{cm}\) and radius of each plane surface is \(3.5\ \text{cm}\); its centre is at O. Find the electric fluxes \(\phi_1,\ \phi_2\) and \(\phi_3\) linked with each plane and the curved surface respectively. What will be the net charge inside the cylinder?
OR
Option 2: Obtain the formula for the capacitance of a parallel plate capacitor when a dielectric medium is partially filled in between its plates.

Hint:

Gauss's law: closed-surface flux \(=q/\varepsilon_0\). Only the two flat faces catch flux (curved surface gives zero); add them and multiply by \(\varepsilon_0\). For the OR part, treat the gap as vacuum plus dielectric layers in series.

Answer 1

Option 1: Gauss's law and cylinder flux (compact approach)

Step 1: Idea. Gauss's law says the closed-surface flux \(\phi=q_{enc}/\varepsilon_0\). Here the field is axial (\(\hat i\) direction), so only the two flat end-caps of the cylinder can catch flux; the curved wall runs parallel to the field lines.

Step 2: End-cap area. \(r=0.035\) m, so \(A=\pi r^2=\pi(0.035)^2=3.85\times10^{-3}\,\text{m}^2\).

Step 3: Both caps point outward with the field. Because \(\vec E\) points \(+\hat i\) for \(x>0\) and \(-\hat i\) for \(x<0\), at each end-cap the field is directed out of the cylinder, parallel to that cap's outward normal. Hence each contributes a positive flux of equal size:
\(\phi_1=\phi_2=EA=100\times3.85\times10^{-3}=0.385\ \text{N m}^2/\text{C}.\)

Step 4: Curved wall. Field is tangential to the curved wall (normal is radial), so \(\phi_3=0\).

Step 5: Net charge. Total outgoing flux \(=2EA=0.770\ \text{N m}^2/\text{C}\), so
\(q_{enc}=\varepsilon_0(2EA)=8.85\times10^{-12}\times0.770=6.8\times10^{-12}\ \text{C}.\)
The positive result means a small net positive charge sits inside the cylinder.
\[\boxed{\phi_1=\phi_2=0.385,\ \phi_3=0,\ q\approx6.8\times10^{-12}\ \text{C}}\]

Option 2: Partially filled capacitor (series-capacitor method)

Step 1: Split into two capacitors in series. The gap of thickness \(d\) is treated as a vacuum layer of thickness \((d-t)\) and a dielectric layer of thickness \(t\) stacked on top of each other, sharing the same charge \(Q\); that is exactly two capacitors in series.

Step 2: Capacitance of each layer.
Vacuum layer: \(C_1=\dfrac{\varepsilon_0 A}{d-t}\).
Dielectric layer: \(C_2=\dfrac{K\varepsilon_0 A}{t}\).

Step 3: Series combination.
\(\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}=\dfrac{d-t}{\varepsilon_0 A}+\dfrac{t}{K\varepsilon_0 A}=\dfrac{1}{\varepsilon_0 A}\left[(d-t)+\dfrac{t}{K}\right].\)

Step 4: Invert.
\[\boxed{C=\frac{\varepsilon_0 A}{d-t+\dfrac{t}{K}}}\]
This matches the field-integration result and reduces correctly to \(\varepsilon_0A/d\) for \(t=0\) and \(K\varepsilon_0A/d\) for \(t=d\).