Question

Q.5. The electric field intensity at a
point on the surface of a long, charge
cylinder in air is 100V/m. Find the
surface charge density on the
cylinder.

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Answer 1

Explanation:

we use the formula of surface charge density

Answer 2

$8.85 X 10^{-10} cm^{2}$ is surface charge density on the cylinder.

Find the surface charge density on the cylinder :

Given

• Charge cylinder in air ⇒ E = 100 V/m,

• ε$_{0}$ = $8.85 X 10-12 C^{2} /N.m^{2}$

•  k = 1 (for air)  

• Let's just be the cylinder's surfaces charge density..

• E =  σ/ ε$_{0}$k  

• σ= Eε$_{0}$k
• $= 100 X 8.85 X 10-2 X 1$  

• = $8.85 X 10^{-10} cm^{2}$

Solution :

• $8.85 X 10^{-10} cm^{2}$ is surface charge density on the cylinder.