Answer :
Answer:
70509.8039216 N/C
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
q = Charge = 2.00 µC
l = Length of filament = 5.1 m
r = Radius of cylinder = 10 cm
[tex]\lambda=\dfrac{q}{l}[/tex]
Electric field is given by
[tex]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[/tex]
The electric field at the surface of the cylinder is 70509.8039216 N/C
Answer:
[tex]E=7368.2844\ N.C^{-1}[/tex]
Explanation:
Given:
- length of the charged filament, [tex]l=0.051\ m[/tex]
- total charge on the filament, [tex]Q=2\times 10^{-6}\ C[/tex]
- length of the cardboard cylinder, [tex]L=0.048\ m[/tex]
- radius of the cylinder, [tex]r=0.1\ m[/tex]
Now form the Gauss's law the electric field on the cylindrical surface:
[tex]E=\frac{Q}{L\times 2\pi.r.\epsilon_0}[/tex] since the charge enclosed by the surface is only upto the length L.
[tex]E=\frac{2\times 10^{-6}}{0.048\times 2\times \pi\times0.1\times 9\times 10^{-9}}[/tex]
[tex]E=7368.2844\ N.C^{-1}[/tex]
is the electric field at the surface of the given cylinder