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Physics 19 Online
OpenStudy (anonymous):

Charged particles repelling + attracting?(Theoretical question!) Ok, Now imagine we have two identical charges (-) with the same size and electrical field strength. They well be repelled Equally, right? However! If we had Identical charges of different size!(Imagine) One larger with a stronger electrical field and another weaker one with a weaker electrical field! The weaker/smaller charge will be repelled far more greater! It would feel a significant force of repulsion from the stronger/large charge! Vice versa with the idea of attraction! I'm I getting this right?

OpenStudy (anonymous):

The whole idea is to illustrate that the LARGER a charge with a strong force would create a powerful repulsion/attraction force on a weaker force of a smaller charge. Makes sense? Imagine having two balls one 20 inch and the other is a 5 inch, push them together close with the same force what would happen? my logic is the small ball (5 inch) would be pushed back GREATLY, same concept is applied with the charges. Hope,

OpenStudy (anonymous):

@experimentX @Jemurray3

OpenStudy (anonymous):

@waterineyes

OpenStudy (anonymous):

Need you're help with this idea! :) Thanks!

OpenStudy (experimentx):

the apple pulls the earth with same force as earth pulls the apple (newton's third law). it seems that apple falls to the Earth ... (newton's second law)

OpenStudy (xishem):

Yes. Newton's third law. When two objects exert forces upon one another, the forces are equal in magnitude and opposite in direction.

OpenStudy (xishem):

You talk about charges of different size. Yes, the smaller (assuming smaller means less mass) charge will experience a greater acceleration because of Newton's second law (F=ma), but the force it experiences will be the same.

OpenStudy (anonymous):

@Xishem True. However, what if the two object where of the same size? Understandable that there will be an equal force. What if there were not of the same size?

OpenStudy (anonymous):

Ow! So "Technically" I'm right :D ?

OpenStudy (xishem):

Technically, no.

OpenStudy (experimentx):

F1 = F2 ma = MA a/A = m/M ... for same force, they have different acceleration, because of their masses

OpenStudy (xishem):

The force experienced by the two charges is the same.

OpenStudy (xishem):

Do not confuse acceleration and force.

OpenStudy (anonymous):

Ow I see. You made sense there.

OpenStudy (anonymous):

so there force is equal no matter the mass. However, their acceleration is not?*

OpenStudy (anonymous):

their*

OpenStudy (xishem):

The force is equal based on Newton's third law, yes. The accelerations will only be the same (in magnitude!) if the two masses are equal. Remember, though, that the accelerations will always be in opposite directions of eachother.

OpenStudy (anonymous):

Confusing matter! However, interesting.

OpenStudy (anonymous):

So lets say I was running super fast! And another person x2 my size ran at the same speed we collided! What would be the result? I accelerate faster back then he will? However, there is the same force? But I accelerated faster lets say x2 before that would mean F= m x 2(a) ?

OpenStudy (xishem):

The force exerted would be the same on both of you. \[F=(1/2)m*2a\]the 1/2 and 2 cancel and you get the same force.

OpenStudy (xishem):

It's better if you say mass than size, as size isn't necessarily proportional to mass. I'll assume you understand that discrepancy.

OpenStudy (anonymous):

Perfect. Now we've forgot an important factor that could change you're answer! The reason why I used a charge... Its because they have electrical forces! Before they are collided.

OpenStudy (anonymous):

Most object before they collide do not have a force present. Only after they collide they create a force that is equal.

OpenStudy (xishem):

Electric charge produces a force as well, so F=ma still applies.\[F_e=k_e \frac{|q_1||q_2|}{r^2}\]

OpenStudy (xishem):

F=ma doesn't just apply to forces of collision -- it applies to all forces. And Newton's third law applies to all forces between two objects.

OpenStudy (anonymous):

Fair enough! Thanks @Xishem .

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