Question

If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to the new value of the electric field at that same point?
a) 1:2
b) 1:3
c) 1:6
d) 1:9

Answer 1

Answer:

b)1 :3

Explanation:

Lets that

The value of a positive charge = q

As we know that electric filed on a point charge given as

$E=\dfrac{Kq}{r^2}$

Where ,K=Constant

q=Charge ,r=Distance

If the value of the charge gets tripled ,q'= 3 q

Then electric filed E'

$E'=\dfrac{Kq'}{r^2}$

$E'=\dfrac{3Kq}{r^2}$

E' = 3 E

Therefore we can say that

$\dfrac{E}{E'}==\dfrac{1}{3}$

therefore the answer will be --

b)1 :3