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A hollow charged conductor has a tiny ho...

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hold is `(sigma)/(2pi epsilon_(0))hatn` where `hatn` is the unit vector in the outward normal direction, and a is the surface charge density near the hole.

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As shown in figure, assume that the hole on the conducting surface is filled by conductor. According to Gauss.s law, at A on surface,
`oint vecE.dvecS =q/epsilon_(0)`
`EdS cos0^(@) = (sigmadS)/epsilon_(0) [therefore sigma = q/(ds)]`
`therefore E.dS = (sigmadS)/(epsilon_(0))`
`therefore vecE = sigma/(epsilon_(0)).hatn`
Which is electric field outside the conductor. Above electric field can be written as,
`vecE = vecE_(1) + vecE_(2)`..........(2)
where `vecE_(1)` = electric field at point `P_0` just outside the conductor (adjacent to point P on the surface) due to charge at point P and `vecE_(2)`= electric field at point P (or at `P_0`) flue to remaining charges on the given charged hollow conductor.
When hollow charged conductor does not have any hole at point P on the surface we have `vecE_(1)` and `E_2` pointing along `hatn` and so from equation (2), we can write,
`E = E_(1) + E_(2) = sigma/(epsilon_(0))`...........(3)
Now consider point `P_(p)` , adjacent to point P and just inside the conductor in figure (a), where resultant electric field `vecE` is `vec0`. Hence,
`vecE = vecE_(1) + vecE_(2) = vec0`
`therefore vecE_(1) = - vecE_(2)`
`therefore vecE_(1) = E_(2)` (Taking magnitude).......(4)
From eqn. (3) and (4)
`2E_(2) = sigma/(epsilon_(0))`
`therefore E_(2) = sigma/(2epsilon_(0))`..........(5)
Now, when we have a hole at point P electric field at point P,- in the hole due to remaining charges will be,
`E_(2) = sigma/(2epsilon_(0))hatn`.............(6)
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Explore conceptually related problems

A hollow charged conductor has a tiny hole cut in to its surface show that the electric field in the hole is n where n is the charge density near the hole

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by (E_(2)-E_(1)).n = sigma/epsilon_(0) where hatn is a unit vector normal to the surface at a point and sigma is the surface charge density at that point. (The direction of hatn is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is sigma hatn //epsilon_(0) . (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.]

Knowledge Check

  • lf the electric field around a surface is given by |vecE|=(Q)/(E_(0)|vecA|) where vecA is the normal area of surface and Q_("in") is the charge enclosed by the surface. This relation of Gauss's law is valid when

    A
    the surface is equipotential.
    B
    the magnitude of the electric field is constant.
    C
    the magnitude of the electric field is constant and the surface is equipotential.
    D
    for all the Gaussian surfaces.

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