Question

intensity of electric field at P is E. If charge is increased by 50% then intensity of field at P changed by

Answer 1

Explanation:

change in E=150%E-100%E=50%E...is your answer my friend

Answer 2

Answer:

Explanation:

Let the charge be q coulomb.

New charge=q+(50/100)*q=3q/2

Now, electric field=kq/(r*r)where k is constant,r is the distance

For new charge, electric field becomes 1.5 times of old field.

So,new field gets increased by 50%.

Hope you understood