# Two point charges Q and 4Q arr 12 cm apart.Sketch the lines of force and calculate the distancr of neutral point from 4Q charge

Manas
13 Points
5 years ago
Dear Shrawni,
I too study in class 12 so my answer can be problematic...but I tried to make it as easy as possible.
So,
Think the neutral charge is kept on the line joining Q and 4Q and x distance from 4Q.
If you know the the electric field by point charge then you too know about the impact of it.
Both the charges will apply their electric field on charge to attract it but neither of them will be able to attract the neutral.
You must be thinking why I didn’t apply coulomb’s law:- Since the charge of neutral particle is zero.
To remain in equilibrium condition the value of electric fields from both the given charges must be equal.So,
E by 4Q=k(4Q)/x^2;
E by Q=kQ/(12-x)^2;
after equating each of them
x=8 cm.
Kartik Tawalare
30 Points
5 years ago
8 cm from 4Q charge
As 4Q/x2 =Q/(12-x)2
4(12-x)2=x2
(12-x)2 = x2/22
(12-x) = x/2
-x = x/2-12
-x = x-24/2
X-24 = -2X
-24 = -2X-X
-24 = -3X
X= 24/3
X=8