Find the total electric flux through the cylinder.

Question

A uniformly charged, straight filament 7.10 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder 4.30 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder.

Using reasonable approximations, find the total electric flux through the cylinder.

Answer 1

I make the accompanying two suppositions:

1) The electrical field through the chamber is opposite to the outer layer of the chamber.

2) The items are in a vacuum.

I believe that the educator has these presumptions on mind when he/she discusses "sensible estimation".

I suspect as much since we need more data to tackle the issue without these suppositions.

All things considered we can apply the Gauss' Regulation

Φ = Q/ε0

where

Φ is the motion we are to ascertain

Q is the charge causing the motion

ε0 = 8.85×10-12 C2kg-1m-3s2 is the permittivity of vacuum.

We can apply the Gauss' Regulation for the accompanying explanation.

Envision that the chamber has tops at the two finishes,

which would transform the chamber into a shut surface.

Then the Gauss' Regulation is material, where

Q = 2.00 µC × 4.30 cm/7.10 m

which is the part of the charge inside the chamber.

Then we notice that by suspicion 1) the transition through the covers is 0,

so Φ is the motion through the side alone.

Connecting numbers

Φ = (2.00 µC × 4.30 cm/7.10 m)/8.85×10-12 C2kg-1m-3s2 ≈ 1360 kgm³/Cs²

Answer 2

I make the following two assumptions:

1) The electrical field through the cylinder is perpendicular to the surface of the cylinder.

2) The objects are in a vacuum.

I think that the teacher has these assumptions on mind when he/she talks about "reasonable approximation".

I think so because we do not have enough information to solve the problem without these assumptions.

In that case we can apply the Gauss's Law

Φ = Q/ε0

where

Φ is the flux we are to calculate

Q is the charge causing the flux

ε0 = 8.85×10^-12 C²kg^-1m^-3s² is the permittivity of vacuum.

We can apply the Gauss's Law for the following reason.

Imagine that the cylinder has lids at both ends,

which would turn the cylinder into a closed surface.

Then the Gauss's Law is applicable, where

Q = 2.00 µC × 4.30 cm / 7.10 m

which is the portion of the charge inside the cylinder.

Then we notice that by assumption 1) the flux through the lids is 0,

so Φ is the flux through the side alone.

Plugging in numbers

Φ = (2.00 µC × 4.30 cm / 7.10 m) / 8.85×10^-12 C²kg^-1m^-3s² ≈ 1360 kgm³/Cs²