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A negative point charge placed at the point A is
- a)In stable equilibrium along x-axis
- b)In unstable equilibrium along y-axis
- c)in stable equilibrium along y-axis
- d)in unstable equilibrium along x-axis
Correct answer is option 'C,D'. Can you explain this answer?
Verified Answer
A negative point charge placed at the point A is a)In stable e...
If the potential energy of the system is minimum, it will be stable equilibrium, i.e , d2U/dx2>0 and when potential energy is maximum then it will be unstable equilibrium, i.e, d2U/dx2<0.
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
Most Upvoted Answer
A negative point charge placed at the point A is a)In stable e...
If the potential energy of the system is minimum, it will be stable equilibrium, i.e , d2U/dx2>0 and when potential energy is maximum then it will be unstable equilibrium, i.e, d2U/dx2<0.
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
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Community Answer
A negative point charge placed at the point A is a)In stable e...
If you displace it a bit towards left on the x-axis the force of attraction by the left charge increases(bcoz distance decline). Thus it goes and attaches to that charge. i. e. Unstable equilibrium
On the y axis though, on displacement the change of distance is equal from both the charges so force applied to the - ve charge is equal from both sides,therefore it gains its original position soon after displaced. i. e stable equilibrium
On the y axis though, on displacement the change of distance is equal from both the charges so force applied to the - ve charge is equal from both sides,therefore it gains its original position soon after displaced. i. e stable equilibrium
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