Question

║⊕QUESTION⊕║
Physics tells us observations can't be predicted absolutely.
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CLASS 12
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A cylinder of radius 5 cm and length 20 cm is bisected perpendicular to its by a plane sheet of a large area which is uniformly charged to a surface charge density 8.854 * 10⁻¹⁰ cm². Calculate the flux passing through a cylinder.

Answer 1

Answer:

6.28 × 10⁴ N m² / C

Explanation:

Given :

Charge density = 8.854 × 10⁻¹⁰ cm²

Radius of cylinder = 5 cm

Length = 20 cm

Area inside cylinder = 2 π r l

= > 2 × 5 × 20 π cm²

= > 200 π cm²

We know :

q = Area × charge density

q = 8.854 × 10⁻¹⁰ × 200 π

We are asked to calculate flux :

We also know :

Flux = q / ε

Flux = (  8.854 × 10⁻¹⁰ × 200 π ) / 8.854 × 10⁻¹²

Flux = π × 2 × 10⁻⁸ / 10⁻¹²

Flux =  2 π 10⁴ N m² / C

Flux = 6.28 × 10⁴ N m² / C

Therefore , flux passing through a cylinder is 6.28 × 10⁴ N m² / C

Answer 2

Question

A cylinder of radius 5 cm and length 20 cm is bisected perpendicular to its by a plane sheet of a large area which is uniformly charged to a surface charge density 8.854 * 10⁻¹⁰ cm². Calculate the flux passing through a cylinder.

Solution

Electric Flux would be 62.8 × 10^-2 Nm²/C

Given

• Length of the Cylinder,L = 20 cm = 20 × 10^-2 m

• Radius of Cross Section,R = 5 cm = 5 × 10^-2 m

• Surface Charge Density = 8.854 × 10^ -10 C/m²

To finD

Flux through the cylinder

$\rule{300}{2}$

A cylindrical guassian surface is bisected perpendicularly by a uniformly charged plane sheet. Firstly,we need to find the electric field in the guassian surface.

Electric Field would be given as :

$\sf \: E = \dfrac{ \sigma}{2 \: \epsilon_o \: } \\ \\ \longrightarrow \: \sf \: E = \dfrac{8.854 \times {10}^{ - 10} }{2 \times 8.854 \times {10}^{ - 12} } \\ \\ \longrightarrow \: \sf \: E = 0.5 \times {10}^{2} \\ \\ \longrightarrow \: \boxed{\boxed{ \sf{E = 50 \: N {C}^{ - 1} }}}$

Now,

$\huge{\boxed{\boxed{ \sf \: \phi \: = 2EA}}}$

$\sf{Here} \begin{cases} \sf{\phi \longrightarrow Flux} \\ \sf{E \longrightarrow Electric \ Field } \\ \sf{A \longrightarrow Area \ of \ the Surface } \\ \sf{\sigma \longrightarrow Surface \ Charge \ Density} \end{cases}$

We would need to find the total surface area of cylinder

$\sf \: A = 2\pi \: rl \\ \\ \leadsto \: \sf \: A = 2(3.14) \times 20 \times {10}^{ - 2} \times 5 \times {10}^{ - 2} \\ \\ \leadsto \: \sf \: A = 628 \times {10}^{ - 5} \\ \\ \leadsto \: \underline{ \boxed{ \sf \: A = 62.8 \: \times {10}^{ - 4} \: \: {m}^{2}}}$

$\rule{300}{2}$

Electric Flux and Electric Field strength would depend upon the number of field lines in a given area. If the surface is uniformly charged,there would be equal number of field lines everywhere. So, Electric Flux is the same throughout the given guassian surface

$\rule{300}{2}$

Substituting the values,we get :

$\sf \phi \: = 2 \times 50 \times 62.8 \times {10}^{ - 4} \\ \\ \longmapsto \: \sf \phi = {10}^{2} \times {10}^{ - 4} \times 62.8 \\ \\ \longmapsto \: \sf \phi \: = 62.8 \times {10}^{ - 2} \: \: \: N {m}^{2} {C}^{ - 1}$

$\rule{300}{2}$

$\rule{300}{2}$