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For a uniformly charged sphere electric ...

For a uniformly charged sphere electric field `vecE` inside the sphere is given as `vecE=(rho)/(3epsilon_(0))(vecr)`, where `rho` is volume charge density and `vecr` is the position vector of the point with respective to centre For a given charge distribtution if the electric field at a point (P) is `vecE_(1)`, and now some part of charge distribution is removed to form a cavity, electric field at same point (P) only due to removed part is `vecE_(2)`. Then finally electric field at point P is `vecE_(3)=vecE_(1)-vecE_(2)`.
There is a uniformly charged sphere of volume charge density `rho` and radius R with a cavity of radius `(R//2)` as shown in the diagram.

The electric field at C

A

`(rho)/(3epsilon_(0))((R)/(2))`

B

`(rho)/(3epsilon_(0))(R)`

C

zero

D

None of these

Text Solution

Verified by Experts

Electric field at C due to whole sphere
`vecE_(1)=(rho)/(3epsilon_(0))(vecr_(c))`
Electric field at C due to only sphere of cavity
`vecE_(2)=(rho)/(3epsilon_(0))(0)=0`
Net field at `CvecE_(3)=vecE_(1)-vecE_(2)=(rho)/(3epsilon_(0))(vecr_(c))=(rho)/(3epsilon_(0))((R)/(2))`
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Knowledge Check

  • The electric field due to a uniformly charged sphere is maximum at :

    A
    centre of sphere
    B
    inside the surface
    C
    at the surface
    D
    at infinite distance from the centre

  • A hollow sphere of copper is positively charged. Then the electric field inside the sphere is

    A
    the same as the field at the surface
    B
    greater than the field at the surface
    C
    ess than the field at the surface but not zero
    D
    zero

  • Electric field intensity at point P due to charge distributed over a sphere is:

    A
    `E = (1)/( 4 pi epsi _(0)) (q)/(r)`
    B
    `E = ( sigma R ^(2))/( epsi _(0)r ^(2))`
    C
    `E = (1)/(4 pi epsi _(0)) (q)/(r)`
    D
    `E =0`

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