Understanding on potential near a charge

  • Thread starter Thread starter rajeshmarndi
  • Start date Start date
  • Tags Tags
    Charge Potential
AI Thread Summary
The discussion centers on understanding electric potential differences in relation to point charges, specifically the signs of potential differences in two figures. It is established that the potential difference Vp - VQ in figure 1 is positive, indicating that point P has a higher potential than Q, while in figure 2, VQ - VP is also positive, suggesting VQ is greater than VP, which confuses some participants. The concept that electric potential is defined as the work done per unit positive charge is emphasized, clarifying that points near a negative charge have lower potential compared to those near a positive charge. Additionally, the role of test charges, whether positive or negative, is discussed, highlighting that potential is a scalar quantity and the direction of work done is crucial in understanding energy interactions in electric fields. The conversation concludes with a reminder that the negative sign in potential does not imply direction but rather indicates a decrease in potential energy in the context of electric fields.
rajeshmarndi
Messages
319
Reaction score
0
From the fig 1 & 2 in the attachment.

i) what is the sign of potential difference in fig.1 Vp – VQ
ii) And in fig.2 VQ – VP

For simplicity, let's take value at P,Q and R as in the figure(attach) for a unit +ve charge.

Both the answer are positive. I can understand i) is positive but not ii), as that means VQ > VP. I know this is what we get when we use the formulae V = Q/4ε°r. But my understanding near the source charge, the potential to do work should always be greater i.e Vp > VQ .

Electric potential at a place is defined as the potential of a unit +ve charge to do work when placed at that point and therefore point near the source would always have greater potential.What is wrong on my understanding?

Also, can't we take unit -ve charge as test charge, since it too has the same potential to do work as the unit +ve charge, only direction of work done is opposite.Thanks.
 

Attachments

  • potential.png
    potential.png
    3.3 KB · Views: 518
Physics news on Phys.org
rajeshmarndi said:
From the fig 1 & 2 in the attachment.
What is this supposed to be showing us? I am reading these as lines of charges without units for charge or distance given. I don't understand what the arrow is supposed to represent.
i) what is the sign of potential difference in fig.1 Vp – VQ
ii) And in fig.2 VQ – VP
If I take these points to be the locations of point charges than the PDs will be undefined.
For simplicity, let's take value at P,Q and R as in the figure(attach) for a unit +ve charge.

Both the answer are positive. I can understand i) is positive but not ii), as that means VQ > VP. I know this is what we get when we use the formulae V = Q/4ε°r. But my understanding near the source charge, the potential to do work should always be greater i.e Vp > VQ .
A +ve charge is attracted to a negative charge so you have to do work to take it away. Thus points close to a negative charge have a low potential compared with points a similar distance from the same strength positive charge.

The sign of the potential depends on your reference point.

Also, can't we take unit -ve charge as test charge, since it too has the same potential to do work as the unit +ve charge, only direction of work done is opposite.
Yes, that's quite valid ... the physics works out the same you just have to pay close attention to the signs. The shape of the potential vs space graph will just be upside down compared with the convention.
 
What is this supposed to be showing us? I am reading these as lines of charges without units for charge or distance given. I don't understand what the arrow is supposed to represent.
these arrows are line of electric field of +ve and -ve charge Q.
If I take these points to be the locations of point charges than the PDs will be undefined.
these are points where its potentials are Vp and Vq.
A +ve charge is attracted to a negative charge so you have to do work to take it away. Thus points close to a negative charge have a low potential compared with points a similar distance from the same strength positive charge.
This is where I am confused. Isn't the same work is done by the test charge if placed at that same point. That is same potential magnitude at same distance from the source. Only direction changes but since potential is a scalar quantity direction shouldn't matter.
 
Last edited:
these arrows are line of electric field of +ve and -ve charge Q.
Ah - so there is one charge indicated by the circles with the + and - sign respectively. Note: the letter Q is commonly used to denote a charge.

This is where I am confused. Isn't the same work is done by the test charge if placed at that same point. That is same potential magnitude at same distance from the source. Only direction changes but since potential is a scalar quantity direction shouldn't matter.

In one case work is done on the test charge and in the other work is done by the test charge. Direction matters because work is determined by the change in potential difference, not the magnitude of the potential.
 
I'm little confused with work done still.

Like when you say work is done ON THE TEST CHARGE - is this work done on the test charge BY AN EXTERNAL AGENT in pushing it against the electric field.

And work done BY THE TEST CHARGE - is this work done on the test charge DUE TO SOURCE CHARGE i.e either repellsion or attraction.
 
I think you need to re-examine the relationship between work and energy then revisit the electrostatic case.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elewor.html

The potential is defined as the work (per unit charge) done against the electric field. Thus changes with the field - in the direction the test charge wants to go by itself - would show up as negative.

Note: scalar's can be negative numbers. The negative sign does not have to mean direction.
 
This is from Griffiths' Electrodynamics, 3rd edition, page 352. I am trying to calculate the divergence of the Maxwell stress tensor. The tensor is given as ##T_{ij} =\epsilon_0 (E_iE_j-\frac 1 2 \delta_{ij} E^2)+\frac 1 {\mu_0}(B_iB_j-\frac 1 2 \delta_{ij} B^2)##. To make things easier, I just want to focus on the part with the electrical field, i.e. I want to find the divergence of ##E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2##. In matrix form, this tensor should look like this...
Thread 'Applying the Gauss (1835) formula for force between 2 parallel DC currents'
Please can anyone either:- (1) point me to a derivation of the perpendicular force (Fy) between two very long parallel wires carrying steady currents utilising the formula of Gauss for the force F along the line r between 2 charges? Or alternatively (2) point out where I have gone wrong in my method? I am having problems with calculating the direction and magnitude of the force as expected from modern (Biot-Savart-Maxwell-Lorentz) formula. Here is my method and results so far:- This...
Back
Top