MCAT physics

[DiamondBar]

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Heyyy I have 2 questions:

1) An electric dipole consists of a pair of equal but opposite charges
+Q and -Q separated by distance d. What is the electric potential at the point (call it P) that's midway between these source charges?

I understand that the answer is 0 (because electric potential is scalar form of understanding charges). I don't understand why electric FIELD is NOT 0. I can sort of picture why (how there is an arrow going straight from +Q to -Q, but equation-wise still seems to add up to 0. If I'm understanding this correctly, Electric field is represented by vectors, and is not scalar, but I'm confused. Can you please help me understand this?

2)
I. Two positive charges, of different masses, placed at d apart.

II. Two negative charges, of equal madsses, placed at d apart.

III. One positive Charge and one negative charge, of equal masses, placed at a distance d apart.

Correct me if I'm wrong, but I and II demonstrate action reaction pair because Q1 is acting an electric force Q2, which Q2 is also doing to Q1 (F=qE)

but I have NO idea why III can be an action reaction pair.

 

[Rabolisk]

1. Your intuition is correct. Mathematically, the magnitudes are equal, and they point in the same direction, so they do not cancel out to 0. Remember that electric field lines go away from positive charge and toward negative charge. If you post your math, I can point out any mistakes.

2. The only difference between 3 and the others is that the force is attractive rather than repulsive. Gravity is the same way. Remember that all forces have a reaction.

 

[DiamondBar]

1. Your intuition is correct. Mathematically, the magnitudes are equal, and they point in the same direction, so they do not cancel out to 0. Remember that electric field lines go away from positive charge and toward negative charge. If you post your math, I can point out any mistakes.

2. The only difference between 3 and the others is that the force is attractive rather than repulsive. Gravity is the same way. Remember that all forces have a reaction.

1) Electric field equation is E= K q/(r^2)

E= K +Q/(r^2) plus K -Q/(r^2) ? It cancels out the same way the potential does. I kept asking my self, "what am I missing???" for hours! I know that the distance is not the problem because even if it were, the r is squared meaning positive and negative direction for vectors wouldn't matter.

2) Then doesn't almost anything in the world has an action-reaction? Maybe I'm not getting the accurate idea of what an "action-reaction" is. If you push a block with a smaller block, I'm obviously exerting a force (small to big block). But there is an opposite force (equal to the previous force) that the bigger block exerts on the smaller block due to the force I'm exerting.

How does this apply to two attractive objects that are not touching but placed at a distance d apart? I'm just confused about this whole thing. I don't even correctly understand how exactly I "know" the first and second is action-pair; they are not touching.


On a side note: Action reaction pair must have two objects with two forces that act on different objects right?



EDIT: I can sort of show this in a picture as well


+Q ---------|--------- -Q
^midpoint
Field here= F | Field here = -F (vector going this way<----)

Field at the midpoint with respect to +Q will be F
Field at the midpoint with respect to -Q will be -F

So adding magnitudes will be F + -(-F)

How does that extra negative sign show up in the equation with sum of the two e=KQ/(r^2) ??
How am I supposed to solve problems like these without picturing them? I prefer to use equations to prove true/false questions, not purely by my intuition. These are the kind of problems that I put the wrong answers to despite my strong intuition.


Sorry if I'm asking juvenile questions; I'm horrible in physics!

 

[Rabolisk]

I will answer the first one when I get back to my apartment and I have access to a laptop. As far as the second question goes, yes, every force has a reaction. It is the definition of newtons third law. If object A exerts a force on object B then there is an equal (magnitude) and opposite (direction) force exerted by object B on A.

 

[DiamondBar]

I will answer the first one when I get back to my apartment and I have access to a laptop. As far as the second question goes, yes, every force has a reaction. It is the definition of newtons third law. If object A exerts a force on object B then there is an equal (magnitude) and opposite (direction) force exerted by object B on A.

So if A is attracted (but not touching) to B, A is exerting force towards B as B is exerting electrical force towards A ?

Then if A is repelled by B, A as electrical forcing going away and so does B, but they are not acting any force against each other, which makes I and II not an action reaction pair 🙁 Ahh!! Sorry my feeble mind cant comprehend, so I need your help!!

I wish I was smarter........

 

[Rabolisk]

1. The math is a bit more difficult because the electric field is a vector, and we have to consider the displacement, not just the scalar distance. Here's a very mathematical and correct way of doing this.

First, if you are not already familiar, the position vector is a vector that points from the origin to a point of interest. So we can say that +Q has a position vector of r1, and that -Q has a position vector of r2, and that the midpoint P has a position vector of r3 (or any other variables). If +Q was located at (2,0,0) on the cartesian coordinate system, its position vector r1 would be 2i, where i is the unit vector in +x direction.

Second, the displacement vector can be defined through position vectors. The displacement vector pointing from +Q to -Q would be (r2-r1).

Finally, the vector form of the equation for electric field at any point p is Ep = kq1*(rp-r1)/|rp-r1|^3. This can be rewritten as Ep = kq1*(rp-r1)hat/|rp-r1|^2 = kq1*rhat/r^2, where r is the displacement between q1 and p. (rp-r1)hat is the unit vector that is in the direction of (rp-r1), r1 is the position vector for q1, and rp the vector for p. If q1 is positive, then the entire expression is positive, which means that Ep = c*rhat, where c is some magnitude and positive. That means that Ep points in the direction of rhat and rhat points from q1 to P. Hence Ep points away from q1. With a negative q1, the opposite is true.

Let's do this problem with some arbitrary coordinates. We'll say that +Q is at (3,0,0), and -Q (5,0,0) and P (4,0,0). So we'll do this problem twice, one for E from +Q and one for E from -Q, and see what happens.

E(+QP) = kQ*rhat/r^2
= kQ*(rp-r1)hat/|rp-r1|^2
= kQ*ihat/1^2
= kQihat.

In another words, the magnitude is kQ, and the direction is to the i direction, or to the right.

E(-QP) = k(-Q)*rhat/r^2
= k(-Q)*(rp-r2)hat/|rp-r2|^2
= k(-Q)*(-ihat)/1^2
= kQihat

You get the same answer, which means same magnitude. Note that -Q cancels out the -ihat. The KEY to this entire problem, and the past 20 minutes of my life was really understanding that the displacement vector from +Q to P and from -Q to P are in opposite direction. You get -ihat because the vector is pointed to the left, from (5,0,0) to (4,0,0). By extension, this shows how electric field lines point towards negative charge and away from positive charge. This quantitative method will work for any problems dealing with electric fields from static charges, but understanding intuitively the direction of the electric field is more useful on the MCAT and in the classroom.

So if A is attracted (but not touching) to B, A is exerting force towards B as B is exerting electrical force towards A ?

Not exactly... If A is attracted to B, that means that B is exerting a force on A, whose direction is towards B. To say that A is exerting force towards B is somewhat wrong, although I think you probably meant to say that A feels the force that is exerted towards B.

If A is repelled by B, then B is repelled by A, and that represents an action-reaction pair. The direction of the force and the physical contact between objects are not relevant. As long as Q1 acts on Q2, then Q2 will do the same (magnitude) and opposite (direction) to Q1. Perhaps what you're confused by is direction. If A is to the left of B, and A is repelled by B, the force on A is to the left. B is repelled by A, and the force on B is to the right, which represent opposite directions.

 

[DiamondBar]

Thank you so much! I cant tell you how much I appreciate such a detailed and enlightening answer!!😀