How is linear momentum conserved in circular motion? Consider a system of two charged particles with opposite charge. Let's assume that the  negatively charged particle revolves around the positively charged particle due to the force of attraction, $\vec{F}$, between them.
In this system, the force causing the negatively charged particle to accelerate is internal, but it changes the direction of velocity, $\vec{v}$ of the negatively charged particle. This means that the linear momentum of the negatively charged particle, $\vec{p} = m\,\vec{v}$, is also changing and is not constant. 
Is this a violation of the law of conservation of momentum?
 A: All conservation laws work for isolated systems. The momentum is conserved for two isolated particles revolving around each other. In your example the change in linear momentum of one particle is taken up by an opposite change in the other.
A: Suppose the system under consideration is the two charges and there are no external forces.  
The force on the positive charge due to the force of attraction of the negative charge $F_{+\,-}$ is equal in magnitude and opposite in direction to the force on the negative charge due to the force of attraction of the positive charge $F_{-\,+}$ as these are a Newton third law pair of internal forces.  
Using Newton's second law $F_{+\,-}= \dfrac{d p_+}{dt}$ and $F_{-\,+}= \dfrac{d p_-}{dt}$ where $p$ is the linear momentum.  
This shows that the magnitude of the (rate of) change in linear momentum for the positive charge is the same as the magnitude of the (rate of) change in linear momentum for the negative charge and as the forces are in opposite directions then so are the respective changes in linear momentum.
The net change in the linear momentum of the system is zero.
It is impossible for one of the charges not to be moving.  
A: Your confusion lies in the way you configured the problem.  Let two charged particles revolve around the center of the system.  It's quite clear in that viewpoint that any change in linear momentum of one particle is matched by a corresponding change of linear momentum in the second particle.  Thus the linear momentum of the whole system remains constant.
However, the way you phrased the question fixes the positively charged particle in place.  Since the positively charged particle is accelerating, you have chosen a non-inertial frame of reference.  The equations of motion are more complex in this frame.  This is the same issue we have with using rotating frames such as ECEF.  We must model the effects such as centripetal accelerations and Coreolis effects.
If you run the math, what you will find is that the pseudoforces associated with your rotating reference frame exactly counter the changes in linear momentum, resulting in momentum being conserved.  Of course doing so requires a lot of extra math.  It's far easier to solve the problem in an inertial frame -- in particular a frame centered on the two particles rather than one particle or the other.
A: First consider the definition of linear momentum of system of particles given below:-
$$\sum_{i=0}^n \vec{p_{i}}=\vec{p_{system}}$$
Now linear momentum of system remains conserved when $\vec{F_{ext}}$ is equal to zero.
In this particular case as the positive charge is fixed it means that system acted by an external force to keep the charge fixed so $\frac{d\vec{p_{system}}}{dt}$ is not equal to zero or the linear momentum of the system is not conserved.
