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 flux through the cylinder

Statement of Gauss's law: The total electric flux through any closed surface equals \(1/\varepsilon_0\) times the net charge enclosed by that surface:
\[\phi=\oint \vec{E}\cdot d\vec{A}=\frac{q_{enc}}{\varepsilon_0}.\]
Step 1: Geometry. The cylinder lies along the X-axis with centre O at the origin. Its two flat (plane) faces are at \(x=+15\ \text{cm}\) and \(x=-15\ \text{cm}\). Radius of each flat face \(r=3.5\ \text{cm}=0.035\ \text{m}\).
Area of a flat face:
\[A=\pi r^2=3.14\times(0.035)^2=3.85\times10^{-3}\ \text{m}^2.\]
Step 2: Flux \(\phi_1\) through the right plane face (x > 0). Here \(\vec{E}=+100\,\hat{i}\) points along \(+x\) and the outward normal also points along \(+x\), so the angle between them is \(0^{\circ}\):
\[\phi_1=EA\cos0^{\circ}=100\times3.85\times10^{-3}=+0.385\ \text{N m}^2/\text{C}.\]
Step 3: Flux \(\phi_2\) through the left plane face (x < 0). Here \(\vec{E}\) is negative, i.e. it points along \(-x\) with magnitude \(100\ \text{N/C}\), and the outward normal of this face also points along \(-x\). The angle between them is again \(0^{\circ}\):
\[\phi_2=EA\cos0^{\circ}=100\times3.85\times10^{-3}=+0.385\ \text{N m}^2/\text{C}.\]
Step 4: Flux \(\phi_3\) through the curved surface. On the curved surface the outward normal is radial (perpendicular to the X-axis), while \(\vec{E}\) is along the X-axis. The angle between them is \(90^{\circ}\):
\[\phi_3=EA'\cos90^{\circ}=0.\]
Step 5: Net flux.
\[\phi=\phi_1+\phi_2+\phi_3=0.385+0.385+0=0.770\ \text{N m}^2/\text{C}.\]
Step 6: Net charge inside (Gauss's law).
\[q_{enc}=\varepsilon_0\,\phi=(8.85\times10^{-12})\times0.770.\]\[q_{enc}=6.81\times10^{-12}\ \text{C}\approx6.8\ \text{pC}.\]
\[\boxed{\phi_1=\phi_2=0.385\ \text{N m}^2/\text{C},\ \phi_3=0,\ q_{enc}\approx6.8\times10^{-12}\ \text{C}}\]

Option 2: Capacitance of a parallel plate capacitor partially filled with a dielectric

Step 1: Set-up. Consider a parallel plate capacitor with plate area \(A\) and plate separation \(d\). A dielectric slab of dielectric constant \(K\) and thickness \(t\ (t<d)\) is inserted between the plates. The remaining gap of thickness \((d-t)\) is vacuum (or air).
Step 2: Field in the vacuum region. If \(\sigma\) is the surface charge density on the plates and \(Q\) the charge, the field in the free (vacuum) region is
\[E_0=\frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0 A}.\]
Step 3: Field inside the dielectric. Inside the dielectric the field is reduced by the factor \(K\):
\[E=\frac{E_0}{K}=\frac{Q}{K\varepsilon_0 A}.\]
Step 4: Potential difference between the plates. The total potential drop is the field times distance in each region:
\[V=E_0(d-t)+E\,t=E_0(d-t)+\frac{E_0}{K}t.\]\[V=\frac{Q}{\varepsilon_0 A}\left[(d-t)+\frac{t}{K}\right].\]
Step 5: Capacitance. Using \(C=\dfrac{Q}{V}\):
\[C=\frac{Q}{\dfrac{Q}{\varepsilon_0 A}\left[(d-t)+\dfrac{t}{K}\right]}=\frac{\varepsilon_0 A}{(d-t)+\dfrac{t}{K}}.\]
\[\boxed{C=\frac{\varepsilon_0 A}{d-t+\dfrac{t}{K}}}\]
Check: if \(t=0\) (no dielectric), \(C=\varepsilon_0 A/d\); if \(t=d\) (fully filled), \(C=K\varepsilon_0 A/d\), as expected.