HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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Solution - Find the Mechanical Force per Unit Area of the Charged Conductor. - HSC Science (Electronics) 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

 

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