Question

Five charges, q each are placed at the corners of a regular pentagon of side ‘a’ (Figure).

(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 corners (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 corners?

Answer 1

(a) (i) The point O, the centre of the pentagon is equidistant from all the charges at the end point of pentagon. Thus, due to symmetry, the electric field due to all the charges is cancelled out. As a result, electric field at O is zero.

(ii) We can write that the vector sum of electric field due to charge A and electric field due to the other four charges at the centre of cube should be zero or, `vecE_A + vecE_("four charges") = 0`

Hence `vecE_("four charges") = - vecE_A` or ⇒ `|vecE_("four charges")| = |vecE_A|`

When charge q is removed from A, net electric field at the centre due to remaining charges `|vecE_("four charge")| = |vecE_A| = 1/(4 piε_0) q/r^2` along OA.

(iii) If charge q at A is replaced by –q, then electric field due to this negative charge

`vecE_(-q) = 1/(4 piε_0) q/r^2` along OA.

HEnce net electric field at the centre

`vecE_("net") = vecE_(-q) + vecE_("four charges") = 1/(4 pi ε_0) q/r^2 + 1/(4 pi ε_0) q/r^2`

`vecE_("net") = 1/(4 pi ε_0) (2q)/r^2` along OA.

(b) If pentagon is replaced by n-sided regular polygon with charge q at each of its comers. Here again, charges are symmetrical about the centre. The net electric field at O would continue to be zero, it doesn’t depend on the number of sides or the number of charges.