Five charges,q each are placed at the comers of a regular pentagon of side a.          

i) What will be the electric field at O, the centre of the pentagon?
(ii)
What will be the electric field at O if the charge from one of the comers (say A) is removed?
(iii)
What will be the electric field at O if the charge q at A is replaced by – q?
(b)
How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its comers?

0 31 Views | Posted 4 months ago
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  • 1 Answer

  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar

    Explanation-  (a)

    (i) The electric field at the center of pentagon is zero because the distance from the center is same.

    (ii) The field through one charge is Kq/r2

    (iii) When one charge is positive and other is negative then net force towards negative charge. So net force is Kq/r2+ Kq/r2= 2Kq/r2

    (b) It doesn't depend upon the number of sides increasing the net electric field is zero.

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