Show that the electric at the surface of a charged conductor is given by -E-0n- where is the charge density is a unit vector normal to the surface in the outward direction-

Electric field= KQR^2=KR^2×σ× 4π R^2 =1R^2×σ× 4π R^2 =σE_0 because it is very near of in same direction n̂ ...

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Standard IX

Chemistry

Molecules and Atomicity

Show that the...

Question

Show that the electric at the surface of a charged conductor is given by $\to E = \frac{σ}{ε_{0}}^n$ where is the charge density is a unit vector normal to the surface in the outward direction.

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Solution

Electric field= $\frac{K Q}{R^{2}} = \frac{K}{R^{2}} \times σ \times 4 π R^{2}$
$= \frac{1}{R^{2}} \times σ \times 4 π R^{2}$
$= \frac{σ}{E_{0}}$
because it is very near of in same direction $^n$

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Q. A nollow charged conductor has a tiny hole cut into its surface. show that the electric fleld in the hole is

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Q. If

σ

is the surface charge density, then electric field at a point near the surface of a charged conductor kept in a dielectric medium of constant

k

is given by

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

Where is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

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Show that the electric at the surface of a charged conductor is given by →E=σε0^n where is the charge density is a unit vector normal to the surface in the outward direction.

Electric field=KQR2=KR2×σ×4πR2=1R2×σ×4πR2=σE0because it is very near of in same direction^n

Show that the electric at the surface of a charged conductor is given by $\to E = \frac{σ}{ε_{0}}^n$ where is the charge density is a unit vector normal to the surface in the outward direction.

Electric field= $\frac{K Q}{R^{2}} = \frac{K}{R^{2}} \times σ \times 4 π R^{2}$
$= \frac{1}{R^{2}} \times σ \times 4 π R^{2}$
$= \frac{σ}{E_{0}}$
because it is very near of in same direction $^n$