Is the statement 'Particle A has twice the charge of particle B' true or false?

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The statement "Particle A has twice the charge of particle B, thus the force particle A exerts on particle B is twice as great as the force particle B exerts on particle A" is false. According to Newton's third law, the forces between two charged particles are equal in magnitude and opposite in direction, regardless of their charges. Therefore, while particle A may have a greater charge, it does not mean it exerts a force that is twice as strong on particle B. The interaction is governed by the principle that both particles exert equal and opposite forces on each other. This highlights the importance of understanding the relationship between charge and force in physics.
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Charge Problem--True/False

Homework Statement


Please explain why this statement is false:

Particle A has twice the charge of particle B. Thus the force particle A exerts on particle B is twice as great as the force particle B exerts on particle A.

Homework Equations


The Attempt at a Solution



I believe it because there will be cancellation thus particle B will not exert any charge on A, but I'm not sure.
 
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btpolk said:

Homework Statement


Please explain why this statement is false:

Particle A has twice the charge of particle B. Thus the force particle A exerts on particle B is twice as great as the force particle B exerts on particle A.

Homework Equations



The Attempt at a Solution



I believe it because there will be cancellation thus particle B will not exert any charge on A, but I'm not sure.
Ask Isaac Newton. See his 3rd law.
 


So am I right by saying one half of particle A's force is being exerted on particle B?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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