A uniformly charged, straight filament 6 m in length has a total positive charge of 3 µC. An uncharged cardboard cylinder 1 cm in length and 5 cm in radius surrounds the filament using the filament as its axis of symmetry, with the filament as the central axis of the cylinder. Find the total electric flux through the cylinder. The permittivity of free space is 8.8542 × 10−12 C 2 /N · m2 . 1. 501.959 2. 4356.29 3. 1452.1 4. 1355.29 5. 564.704 6. 250.979 7. 4065.87 8. 141.176 9. 3049.4 10. 847.056 Answer in units of N · m2 /C. 009 (part 2 of 2) 10.0 points What is the electric field at the surface of the cylinder? 1. 224688.0 2. 22468.8 3. 503302.0 4. 209709.0 5. 161776.0 6. 462216.0 7. 719003.0 8. 171191.0 9. 114127.0 10. 179751.0 Answer in units of V/m

Part A : 5.) = 564.704 N*m²/C

Part B:  10.) = 179751.0 V/m

Explanation:

A) Applying Gauss'Law to the straight filament, using a cylindrical gaussian surface with the filament as the central axis of the surface, assuming that the electric field is normal to the surface (which means that no flux exist through the lids of the cylinder) and is constant at any point of the surface (except the lids where is 0), we can find the electric flux, as follows:

where:

r is the radius of the cylinder = 0.05 m

l is the length of the cylinder = 0.01 m

Qenc, is the net charge on the filament enclosed by the gaussian surface

ε₀ = 8.8542*10⁻¹² C²/N*m²

In order to find the value of Qenc, we need to find first the linear charge density, as follows:

The net charge enclosed by the gaussian surface will be just the product of the linear change density λ times the length of the gaussian surface:

According Gauss ' Law, the net flux through the gaussian surface must be equal to the charge enclosed by the surface, divided by the permittivity of free space (in vacuum or air), as follows:

which is the same as the option 5.

B)  Repeating the equation (1) from above:

we can solve for E, as follows:

which is the same as the option 10 of part B.

i think the answer would be reserve.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards