Show that the electric at the surface of a charged conductor is given

1.	Show that the electric at the surface of a charged conductor is given by E = σ/ε0n cap , where where σis the surface charge density and n cap is a unit vector normal to the surface in the outward direction.
Answer» Answer: Explanation: According to GAUSS's theorem, $\int\limits {E} \, ds$ = q/ε Here, q is the charge of the conductor. Now, σ=q/ds q=σds Hence, E.dscosθ = σds/ε As the unit VECTOR is NORMAL to to the surface in the upward direction. Thus, the angle between the electric FIELD and area vector is 0. E.ds = σds/ε E=σ/ε Along the vector- E=(σ/ε)N cap

Show that the electric at the surface of a charged conductor is given by E = σ/ε0n cap , where where σis the surface charge density and n cap is a unit vector normal to the surface in the outward direction.

Answer»

Answer:

Explanation:

According to GAUSS's theorem,

$\int\limits {E} \, ds$ = q/ε

Here, q is the charge of the conductor.

Now, σ=q/ds

q=σds

Hence, E.dscosθ = σds/ε

As the unit VECTOR is NORMAL to to the surface in the upward direction.

Thus, the angle between the electric FIELD and area vector is 0.

E.ds = σds/ε

E=σ/ε

Along the vector-

E=(σ/ε)N cap

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Show that the electric at the surface of a charged conductor is given by E = σ/ε0n cap , where where σis the surface charge density and n cap is a unit vector normal to the surface in the outward direction.​

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Show that the electric at the surface of a charged conductor is given by E = σ/ε0n cap , where where σis the surface charge density and n cap is a unit vector normal to the surface in the outward direction.