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# Point charges + 4q, -q and +4q are kept on the X-axis at points x = 0, x = a and x = 2a respectively.(a)    Only- q is in stable equilibrium (b)    None of the charges is in equilibrium (c)    All the charges are in unstable equilibrium (d)    All the charges are in stable equilibrium.

Apoorva Arora IIT Roorkee
7 years ago
The charge at x=a is experiencing attractive forces from both the charges 4q
$F_{1}=\frac{k(4q)q}{a^{2}}$and$F_{2}=\frac{k(4q)q}{a^{2}}$which are equal in magnitude and opposite in direction. So, The charge -q is in stable equilibrium.
Now, The charge at x=0 is experiencing two forces
$F_{1}=\frac{k(4q)q}{a^{2}}$due to -q and$F_{2}=\frac{k(4q)(4q)}{(2a)^{2}}=\frac{k(4q)q}{a^{2}}$which are equal in magnitude and opposite in direction.(one is attractive and other is repulsive). It is the same case with the charge 4q at x=2a.
Hence, all the charges are in stable equilibrium.
Thanks and Regards
Apoorva Arora
IIT Roorkee
Rijul
11 Points
3 years ago

#### Explanation:

Let's see what is the force acting on charge q, it is k4q2a2 by both the 4q charges but on opposite direction,so q is at equilibrium.

Again, on each of =4q charge,force acting is +k16q2(2a)2k(4q2a2)=k4q2a2k4q2a2=0

So,they are at equlibrium as well,

Now,if we move charge q a bit to either side,due to decreased distance,the side on to which we pushed,the force acting on that direction will increase (as F1r2),so the charge q will start moving towards that charge as on moving further towards it,the force will increase,so q is in unstable equilibrium.

Now,doing the same thing, for 4q,we can clearly call it is as well in unstable equilibrium,as the magnitude of the two other charges are different,it was made stable by balancing the charge inequality with similar inequality of distance between them.