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2 years ago

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A uniformly charged, straight filament 5.10 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder

4.80 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. (a) Using reasonable approximations, find the electric field at the surface of the cylinder.

2 answers:

Ratling [72]2 years ago

7 0

Answer:

70509.8039216 N/C

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = 2.00 µC

l = Length of filament = 5.1 m

r = Radius of cylinder = 10 cm

$\lambda=\dfrac{q}{l}$

Electric field is given by

$E=\dfrac{2k\lambda}{r}\\\Rightarrow E=\dfrac{2\times 8.99\times 10^9\times \dfrac{2\times 10^{-6}}{5.1}}{10\times 10^{-2}}\\\Rightarrow E=70509.8039216\ N/C$

The electric field at the surface of the cylinder is 70509.8039216 N/C

34kurt2 years ago

4 0

Answer:

$E=7368.2844\ N.C^{-1}$

Explanation:

Given:

length of the charged filament, $l=0.051\ m$

total charge on the filament, $Q=2\times 10^{-6}\ C$

length of the cardboard cylinder, $L=0.048\ m$

radius of the cylinder, $r=0.1\ m$

<u>Now form the Gauss's law the electric field on the cylindrical surface:</u>

$E=\frac{Q}{L\times 2\pi.r.\epsilon_0}$ since the charge enclosed by the surface is only upto the length L.

$E=\frac{2\times 10^{-6}}{0.048\times 2\times \pi\times0.1\times 9\times 10^{-9}}$

$E=7368.2844\ N.C^{-1}$

is the electric field at the surface of the given cylinder

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