Q pack Physics 78

[jeep1010]

If the magnitude of a positive charge is tripled, what is the ratio of the original value of the electric field at a point to the new value of the electric field at the same point?

A- 1:2
B- 1:3
C- 1:6
D- 1:9

I used the formula E = F/q (F=qE). I tripled the q which I figured would make the electric field 1/3 of the original (E = F/3q). So, i reasoned 1:3 as the answer.

The AAMC solution mentions using E = kq/r^2

Is my reasoning wrong? When do you use the formula E = F/q vs E = kq/r^2

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[aldol16]

These are basically two forms of the same equation once you realize that F = k*q1*q2/r^2. When you're using E = F/q, you should be cognizant of what that "q" in the divisor is. Is it q1 or q2? Well, what's the definition of electric field? It's the electric force experienced by a positive test charge. So say q1 is the point charge that you're interested in and q2 is the test charge. Then the electric field experienced by that test charge is E = F/q. Well, which q should we use? The test charge serves to test the strength of the electric force at various points. Thus, it makes sense that the "q" in this equation is E = F/q2 because you're "testing" the field exerted by the point charge that you're interested in by some test charge and so you should normalize based on what magnitude that point charge has.

 

[deleted647690]

These are basically two forms of the same equation once you realize that F = k*q1*q2/r^2. When you're using E = F/q, you should be cognizant of what that "q" in the divisor is. Is it q1 or q2? Well, what's the definition of electric field? It's the electric force experienced by a positive test charge. So say q1 is the point charge that you're interested in and q2 is the test charge. Then the electric field experienced by that test charge is E = F/q. Well, which q should we use? The test charge serves to test the strength of the electric force at various points. Thus, it makes sense that the "q" in this equation is E = F/q2 because you're "testing" the field exerted by the point charge that you're interested in by some test charge and so you should normalize based on what magnitude that point charge has.

So then, assuming the equation E = F /q involves q as the test charge, wouldn't it make sense to use the equation they used, since

E = F /q2 (q2 is test charge)

E = k q1 * q 2 / r^2 (q2)

Test charge cancels, leaving you with E = k*q1 / r^2

If you just used E = F/q, you would be assuming they are changing the test charge, but they are changing the point charge in the problem. Even so, you get the same answer in the end

 

[aldol16]

So then, assuming the equation E = F /q involves q as the test charge, wouldn't it make sense to use the equation they used, since

E = F /q2 (q2 is test charge)

E = k q1 * q 2 / r^2 (q2)

Test charge cancels, leaving you with E = k*q1 / r^2

If you just used E = F/q, you would be assuming they are changing the test charge, but they are changing the point charge in the problem. Even so, you get the same answer in the end

That is what I'm saying. The point charge exerts the electric field. The point charge is a real charge - it is a real physical entity. The test charge is imaginary. It's an imaginary particle with x charge that is experiencing the electric field caused by the point charge. Here's the derivation I'm imagining:

q1 is a point charge exerting some electric field and q2 is a test charge moving through that field. q2 will experience a field of:

E = F/q2

F is equal to k*q1*q2/r^2.

Thus,

E = (k*q1*q2/r^2)/q2

E = k*q1/r^2

Your result is therefore that electric field varies inversely with the square of radius for a point charge q1.