Question

two point charges +4q and +q are placed 30cm apart.At what point on the line joining them the electric field is zero?

Answer 1

Where E=0 this point is called neutral point.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi€
for 4q charge
let at ''a'' distance we get E=0 which is from q charge
so distance from 4q of 'a' point is 30-a
electric field by 4q charge on a is
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.
i hope u got ur answer.
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Answer 2

Dear Student,

◆ Answer -

10 cm from q or 20 cm from 4q

◆ Explaination -

Let intensity of electric field be zero at distance x from charge q. Thus the point will be 30-x away from charge 4q.

For electric field intensity to be equal, electric field due to two points should be equal.

E1 = E2

k.q/x^2 = k.4q/(30-x)^2

(30-x)^2 = 4x^2

30-x = 2x

3x = 30

x = 30/3

x = 10 cm

Thus this point will be 10 cm from +q and 30-10 = 20 cm from charge +4q.

Hope this was helpful...