Electric Potential due to multiple charges

AI Thread Summary
The discussion clarifies the concept of electric potential due to multiple charges, emphasizing that the electric potential (V) is determined by the formula V = kQ/r, where Q is the charge and r is the distance from the charge. It highlights that doubling the charge will double the potential, while increasing the distance will decrease the potential. Importantly, the potential at a point is independent of any test charge present. The relevance of the second charge in determining potential is deemed irrelevant in this context. Understanding these principles is crucial for solving problems related to electric potential.
irivce
Messages
11
Reaction score
0

Homework Statement


bfPCQT1.png



Homework Equations



E = kq1q2/r2


The Attempt at a Solution



Please correct me if I am wrong but I think it is A. Since doubling the charge, doubles the potential as well, while doubling the radius between the charges would just exponentially decrease the electric potential. My question here is if that is the correct equation?
 
Physics news on Phys.org
Be careful. This questions asks you which statement is true about the electric potential - not the electric field. Recall that the electric potential produced by a point charge Q is given by the expression:

##V = k \frac{Q}{r}##

where Q is the charge and r is the distance away from this charge. (This is the source charge).
 
So is it safe to say that the charge from the second point is irrelevant ?
 
That is correct. The electric potential at a point in space is independent of the test charge.

Some light reading from wikipedia for you (I especially recommend the introduction and the section titled electric potential due to a point charge): http://en.wikipedia.org/wiki/Electric_potential
 
  • Like
Likes 1 person
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top