Five charges, q each are placed at the corners of a regular pentagon of side ‘ a ’ as shown. What will be the electric field at O if the charge from one of the corners (say A ) is removed?

When charge q is placed at all corners of regular polygon then due to symmetry electric field E ' at O is zero ∴ E ' = 0 Now charge q is removed from corner A therefore, net charge at A becomes zero. Or we can say that charges at corner A are + q , - q (net charge is zero) We know electric field at O due to other four corner and due to charge + q at corner A is zero. Hence, net electric field E at O is due to charge - q at A . ∴ E = 1 4 πε 0 q r 2 along OA where, r = a 2 sin 36 ° is the radius of pentagon

Five charges, q each are placed at the corners of a regular pentagon of side ‘ a ’ as shown. What will be the electric field at O if the charge q at A is replaced by – q ?

When charge q is placed at all corners of regular polygon then due to symmetry electric field E ' at O is zero ∴ E ' = 0 Now charge q is replaced by - q from corner A . Or we can say that charges at corner A are + q , - 2 q (net charge is same i.e. - q ) We know electric field at O due to other four corner and due to charge + q at corner A is zero. Hence, net electric field E at O is due to charge - 2 q at A . ∴ E = 1 4 πε 0 2 q r 2 along OA where, r = a 2 sin 36 ° is the radius of pentagon