Ncert Exercises & Solutions

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

(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 the pentagon is replaced by an n-sided regular polygon with charge q at each of its corners?

Hint: The distance of each charge from the centre is the same and the electric field is a vector quantity.

Step 1: Apply Coulomb's law to find the net electric field at O.

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

(ii) When charge q is removed a negative charge will develop at A giving an electric field,

E = \frac{q}{4 {πε}_{0} r^{2}}

along with OA.

(iii) If charge q at A is replaced by -q, then two negative charges -2q will develop there. Thus, the value of the electric field

E = \frac{2 q}{4 {πε}_{0} r^{2}}

along with OA.

(b) When the pentagon is replaced by an n-sided regular polygon with charge q at each of its corners, the electric ficid at O would continue to be zero as symmetricity of the charges is due to the regularity of the polygon. It doesn't depend on the number of sides or the number of charges.

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