2
Question
Write definition of electric field intensity.
Obtain an expression for electron force and electric pressure on the surcharge of a charged conductor.
Draw necessary diagram.
Obtain an expression for electron force and electric pressure on the surcharge of a charged conductor.
Draw necessary diagram.
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Solution
Electric field intensity : It is the strength of an electric field at any point. It is equal to the electric force per unit charge experienced by a test charge placed at that point. The unit of it i s volt/meter or Newton/ Coulomb..
Every element of a charged conductor experienced a normal mechanical force. This is the result of repulsive force from similar change present on the rest of the surface of the conductor.
Consider a small element ds on the surface of a charge conductor. if σ is the surface density and the charge carried by the element ds be dq.
Consider a point P just outside the surface near the element ds. The electric intensity at a point P is given by E=σ∈o
The direction of the intensity is normally outwards.
The intensity made up of two components →E1 due to the small change dq present on the element ds.
→E2 due to the remaining charge present on the rest of the surface of the conductor.
Hence at P,→E1+→E2=σ∈o
* Now consider a point Q inside the conductor the electric field {E_l} and {E_2} at this point are oppositely directed. However, electric intensity inside a changed conductor is zero
→E1−→E2=0
→E1=→E2
Substituting for E1 in eq.(3) we have
2E2=σ∈o;E2=σ2∈o
Here E2 is the electric field due to the charge on the rest of coductor. Hence repulsive force experienced by elements ds carrying dq to is given by
→f=→E2.dq
F=σ2∈odq
f=σ2∈o.σ.dq
F=sigma22∈o.dq
The electric pressure on surface ds will be
pressure=forceArea
σ22∈odsds
P=σ22∈oN/M2
Every element of a charged conductor experienced a normal mechanical force. This is the result of repulsive force from similar change present on the rest of the surface of the conductor.
Consider a small element ds on the surface of a charge conductor. if σ is the surface density and the charge carried by the element ds be dq.
Consider a point P just outside the surface near the element ds. The electric intensity at a point P is given by E=σ∈o
The direction of the intensity is normally outwards.
The intensity made up of two components →E1 due to the small change dq present on the element ds.
→E2 due to the remaining charge present on the rest of the surface of the conductor.
Hence at P,→E1+→E2=σ∈o
* Now consider a point Q inside the conductor the electric field {E_l} and {E_2} at this point are oppositely directed. However, electric intensity inside a changed conductor is zero
→E1−→E2=0
→E1=→E2
Substituting for E1 in eq.(3) we have
2E2=σ∈o;E2=σ2∈o
Here E2 is the electric field due to the charge on the rest of coductor. Hence repulsive force experienced by elements ds carrying dq to is given by
→f=→E2.dq
F=σ2∈odq
f=σ2∈o.σ.dq
F=sigma22∈o.dq
The electric pressure on surface ds will be
pressure=forceArea
σ22∈odsds
P=σ22∈oN/M2
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