Advanced Level Electric Flux.

Electricity and Magnetism Level 3

A non-conducting, solid cylinder of infinite length has charge density ρ \rho and radius 5 R . 5R.

If the flux passing through the surface A B C D ABCD shown in figure is 2 θ ρ R 3 ε 0 , \large\ \frac { 2\theta \rho { R }^{ 3 } }{ { \varepsilon }_{ 0 } }, find the value of the positive integer θ . \theta.


The answer is 6.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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3 solutions

Rohit Gupta
Feb 22, 2018

The problem can be solved by using the gauss's law on the closed surface ABCD as shown.

Flux through the flat surface + curved surface = ρ V ϵ 0 \dfrac{\rho V}{\epsilon_0}

To calculate the volume, consider the top view of the cylinder. The volume V of the portion is ( π ( 5 R ) 2 53 × 2 360 − 1 2 8 R × 3 R ) × R \left( \pi {(5R)^2}\frac{{53 \times 2}}{{360}}\, - \,\frac{1}{2}8R \times 3R \right) \times R

Flux through the curved surface = Field times curved surface area = ρ 5 R 2 ϵ 0 × 5 R × 53 × 2 × π 180 × R \dfrac{\rho 5R}{2 \epsilon_0} \times 5R \times 53 \times 2 \times \dfrac{\pi}{180} \times R

Putting the values in the equation we get,

Flux through flat surface ABCD + ρ 5 R 2 ϵ 0 × 5 R × 53 × 2 × π 180 × R \dfrac{\rho 5R}{2 \epsilon_0} \times 5R \times 53 \times 2 \times \dfrac{\pi}{180} \times R = ρ ( π ( 5 R ) 2 53 × 2 360 − 1 2 8 R × 3 R ) × R ϵ 0 \dfrac{\rho \left( \pi {(5R)^2}\frac{{53 \times 2}}{{360}}\, - \,\frac{1}{2}8R \times 3R \right) \times R}{\epsilon_0}

Flux through the flat surface ABCD = − 12 ρ R 3 ϵ 0 -\dfrac{12 \rho R^3}{\epsilon_0}

Negative sign implies that the flux is entering into the closed surface through the flat surface ABCD.

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Steven Chase
Feb 15, 2018

Circle Equation:

x 2 + y 2 = 25 R 2 \large{x^2 + y^2 = 25R^2}

Determine x x when y = 4 R y = 4R :

x 2 + 16 R 2 = 25 R 2 x = 3 R \large{x^2 + 16 R^2 = 25R^2 \\ x = 3R}

Consider a Gaussian cylinder of radius r r and length L L . The enclosed charge is:

Q e = π r 2 L ρ \large{Q_e = \pi \, r^2 L \, \rho}

The surface area of the Gaussian cylinder is:

A = 2 π r L \large{A = 2 \, \pi \, r \, L}

Calculate the electric field magnitude from Gauss's Law:

E m = Q e ϵ 0 A = π r 2 L ρ ϵ 0 2 π r L = ρ r 2 ϵ 0 \large{E_m = \frac{Q_e}{\epsilon_0 \, A} = \frac{\pi \, r^2 L \, \rho}{\epsilon_0 \, 2 \, \pi \, r \, L} = \frac{\rho \, r}{2 \epsilon_0}}

Including a directional unit vector yields individual spatial electric field components:

E x = ρ r 2 ϵ 0 x r = ρ 2 ϵ 0 x E y = ρ r 2 ϵ 0 y r = ρ 2 ϵ 0 y E z = 0 \large{E_x = \frac{\rho \, r}{2 \epsilon_0} \frac{x}{r} = \frac{\rho}{2 \epsilon_0} x \\ E_y = \frac{\rho \, r}{2 \epsilon_0} \frac{y}{r} = \frac{\rho}{2 \epsilon_0} y \\ E_z = 0 }

The surface normal components are:

N x = 1 N y = 0 N z = 0 \large{N_x = 1 \\ N_y = 0 \\ N_z = 0}

Flux integral:

ϕ = ∫ 0 R ∫ − 4 R 4 R ( E x N x + E y N y + E z N z ) d y d z = ∫ 0 R ∫ − 4 R 4 R ( E x N x ) d y d z = ∫ 0 R ∫ − 4 R 4 R ( ρ 2 ϵ 0 x ) d y d z = ( 3 R ρ 2 ϵ 0 ) ∫ 0 R ∫ − 4 R 4 R d y d z ( 3 R ρ 2 ϵ 0 ) 8 R 2 = 12 ρ R 3 ϵ 0 \large{\phi = \int_0^R \int_{-4R}^{4R} (E_x \, N_x + E_y \, N_y + E_z \, N_z) \, dy \, dz \\ = \int_0^R \int_{-4R}^{4R} (E_x \, N_x) \, dy \, dz \\ = \int_0^R \int_{-4R}^{4R} \Big(\frac{\rho}{2 \epsilon_0} x \Big) \, dy \, dz \\ = \Big(\frac{3 R \, \rho}{2 \epsilon_0} \Big) \int_0^R \int_{-4R}^{4R} dy \, dz \\ \Big(\frac{3 R \, \rho}{2 \epsilon_0} \Big) \, 8 R^2 \\ = \frac{12 \, \rho \, R^3}{\epsilon_0}}

3 Helpful 1 Interesting 1 Brilliant 0 Confused

@Steven Chase , SIR I have one doubt.

The figure in the question asks for flux through rectangular part whereas the charge density is spread over the volume.

so for calculating the flux by the formula of q e n c l o s e d e p s i l o n n o t \large\ \frac {q_{enclosed}}{epsilon not} ,how can i calculate the charge enclosed for the planar figure(ABCD)? What will be its volume?

We have charge enclosed given by volume charge density multiplied by volume of surface.

What will be the volume for the surface in the question? Please answer in detail.

Priyanshu Mishra - 3 years, 3 months ago

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The rectangular surface isn't closed, so Gauss's law can't be applied in that way. I used a cylindrical Gaussian surface to find an expression for the electric field. And then I evaluated the flux integral of that field over the open rectangular surface.

Steven Chase - 3 years, 3 months ago

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@Steven Chase , okay i missed the basic condition for gauss' law. Thanks for helping.

By the way, Can you help me in this question

"Blowing Mechanics" by me.

Priyanshu Mishra - 3 years, 3 months ago

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@Priyanshu Mishra – No problem. Yes, that blowing problem looks interesting. I'll give it a try some time in the next day or so.

Steven Chase - 3 years, 3 months ago

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@Steven Chase – @Steven Chase , no problem. Thanks.

Priyanshu Mishra - 3 years, 3 months ago
Priyanshu Mishra
Feb 14, 2018

@Steven Chase, please post the solution @Steven https://brilliant.org/profile/steven-77g4y5/

Steven Chase

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Is the answer : 26 ρ R 3 ε 0 \large\ \frac { 26 \rho { R }^{ 3 } }{ { \varepsilon }_{ 0 } } Or 12 ρ R 3 ε 0 \large\ \frac { 12 \rho { R }^{ 3 } }{ { \varepsilon }_{ 0 } }

A Former Brilliant Member - 3 years, 3 months ago

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