HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
Account
It's free!

User


Login
Create free account


      Forgot password?
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution for Find the Mechanical Force per Unit Area of the Charged Conductor. - HSC Science (Computer Science) 12th Board Exam - Physics

Question

A conductor of any shape, having area 40 cm2 placed in air is uniformly charged with a charge 0* 2µC. Determine the electric intensity at a point just outside its surface. Also, find the mechanical force per unit area of the charged conductor.

[∈0=8.85x10-12 S.I. units]

Solution

Given: Q=0.2 μC=0.2*10-6C

A=40cm2=40*10-4m2

ε0=8.85*10-12SI units

The electric field intensity just outside the surface of a charged conductor of any shape is

`E=sigma/epsi_0=Q/(Aepsi_0)`

∴`E=(0.2*10^-6)/(40*10^-4*8.85*10^-12`

∴`E=5.65*10^6`N/C

Now, the mechanical force per unit area of a conductor is

`f=1/2epsi_0E^2=1/2*8.85*10^-12*(5.65*10^6)^2`

∴f=141.25N/m2

 

  Is there an error in this question or solution?

APPEARS IN

Video TutorialsVIEW ALL [1]

Solution for question: Find the Mechanical Force per Unit Area of the Charged Conductor. concept: Mechanical Force on Unit Area of a Charged Conductor. For the courses HSC Science (Computer Science), HSC Science (Electronics), HSC Science (General)
S