Does equal electric field imply equal potential?

AI Thread Summary
Equal electric field does not imply equal electric potential, as the electric field is the gradient of the electric potential. While the electric field strength may be the same at two points, it does not mean the potential is the same; it only indicates that the derivative of potential is constant. In the given scenario with symmetry, the potential will depend only on the radial distance from the origin, suggesting that points equidistant from the origin are equipotential. The potential difference between two points can be calculated using the integral of the electric field along the path between them. The potential change is defined as V_B - V_A when moving from point A to point B.
fishingspree2
Messages
138
Reaction score
0

Homework Statement


Given the E field
E = 18/R2 R, R is the radial direction.
Find the electric potential between A and B where A is at +2m and B at -4m, both on the z axis.


The Attempt at a Solution


My question is, since E field depends only on R, the distance between the point and the origin, then E field is the same for point B at -4m and point C at +4m. Therefore they are equipotential. Therefore I can compute the potential difference between A at +2m and C at +4m. Is this reasoning correct?
 
Physics news on Phys.org
No, equal electric field does not imply equal electric potential. The electric field is the gradient of the electric potential: E = -∇V, or in one dimension, E = -dV/dr.

So, if the electric field strength and direction are the same at two points, it doesn't mean that V is the same at those two points, it just means that the derivative of V is the same at those two points.

Example: electric field between two large parallel plates of uniform surface charge, is the same everywhere between the two plates, but electric potential increases linearly as you move from one plate to the other.

(The approximately uniform gravitational field close to the surface of the Earth is a totally analogous situation. The gravitational field strength is the same at two points at two different heights, but the gravitational potential is definitely different between those two points.)
 
Hmm...what I said above was the most general answer. HOWEVER in this case, it seems like the problem posesses some symmetry. If dV/dr depends only on r, then V will depend only on r as well. So V is the same at all points that are equidistant from the origin.
 
cepheid said:
Hmm...what I said above was the most general answer. HOWEVER in this case, it seems like the problem posesses some symmetry. If dV/dr depends only on r, then V will depend only on r as well. So V is the same at all points that are equidistant from the origin.
Some questions...
I'm trying to compute V between two general points for the given E field, just to see what it will give and also for the sake of it.
We know that

Vb-Va = -\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}

any idea how I could express d\overrightarrow{l}...?
also, should d\overrightarrow{l} be from A to B or from B to A?
is Vb-Va called the potential difference between A and B or the potential difference between B and A?
 
fishingspree2 said:
Some questions...
I'm trying to compute V between two general points for the given E field, just to see what it will give and also for the sake of it.
We know that

Vb-Va = -\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}

any idea how I could express d\overrightarrow{l}...?
also, should d\overrightarrow{l} be from A to B or from B to A?
is Vb-Va called the potential difference between A and B or the potential difference between B and A?
I found that d\overrightarrow{l}=dR\overrightarrow{R}+Rd\phi \overrightarrow{\phi}+dz\overrightarrow{z}

so
V_{B}-V_{A}-\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}=-\int_{R_{A}}^{R_{B}}\frac{18}{R^{2}}dR=\frac{18}{R_{B}}-\frac{18}{R_{A}}
is that correct? in that case it seems my initial assumption that -4 and 4 are equipotential is correct since it depends only on R which is 4 in both cases.
is that the potential change when you go from B to A or is it the potential change when you go from A to B?
 
fishingspree2 said:
I found that d\overrightarrow{l}=dR\overrightarrow{R}+Rd\phi \overrightarrow{\phi}+dz\overrightarrow{z}

so
V_{B}-V_{A}-\int_{A}^{B}\overrightarrow{E}\cdot d\overrightarrow{l}=-\int_{R_{A}}^{R_{B}}\frac{18}{R^{2}}dR=\frac{18}{R_{B}}-\frac{18}{R_{A}}
is that correct? in that case it seems my initial assumption that -4 and 4 are equipotential is correct since it depends only on R which is 4 in both cases.
is that the potential change when you go from B to A or is it the potential change when you go from A to B?

Obviously V_B - V_A is the potential change when you go from A to B. You're taking the difference between the final value and the initial value to compute the change.
 
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