# Need help in understanding Electric field intensity equation

#### kino007

Joined May 24, 2023
7
Good day,

I've been reading a book called "" . In it he gives 3 equations:

1) $$E=Q/4\pi r^2\epsilon_0$$

2) $$f = \Delta q Q/4\pi r^2\epsilon_0$$

3) $$V = \int_{\infty}^{R} fdr = -Q/4\pi r^2\epsilon_0$$

Last edited:

#### ericgibbs

Joined Jan 29, 2010
18,021
hi kino,
Welcome to AAC.
Do you have a question.?
E

#### kino007

Joined May 24, 2023
7
Good day,

I've been reading a book called "Fast Circuit Boards". In it he gives 3 equations:

1) $$E=Q/4\pi r^2\epsilon_0$$

2) $$f = \Delta q Q/4\pi r^2\epsilon_0$$

3) $$V = \int_{\infty}^{R} fdr = -Q/4\pi r^2\epsilon_0$$

My question is:

If the voltage is defined as the work required to move a unit charge between 2 points in an E field, why wasnt the test charge $$\Delta q$$ not included in the equation for the voltage in eqn 3? Its present for the formula of the force experienced by the test charge when placed in an E field (eqn 2) but when integrating over small distances to determine the voltage it disappears.

I was expecting the formula in number 3 to be: $$V = \int_{\infty}^{R} fdr = -Q\Delta q/4\pi r^2\epsilon_0$$

This may be a very noob question but I'm having trouble visualizing it and would really appreciate any help given.

Jermaine

#### kino007

Joined May 24, 2023
7
sorry wasnt done posting was testing if latex worked... turns out we have 10mins to edit after posting, didnt know that, my bad. See full post above

#### kino007

Joined May 24, 2023
7
Good day,

I've been reading a book called "" . In it he gives 3 equations:

1) $$E=Q/4\pi r^2\epsilon_0$$

2) $$f = \Delta q Q/4\pi r^2\epsilon_0$$

3) $$V = \int_{\infty}^{R} fdr = -Q/4\pi r^2\epsilon_0$$
hi kino,
Welcome to AAC.
Do you have a question.?
E
Yea sorry about the confusion. I made a reply with the full post

#### WBahn

Joined Mar 31, 2012
29,164
The force felt by a charge is proportional to the product of both charges. If the test charge is twice as large, force experienced by it due to other charges will be twice as great. However, voltage is defined so that it is independent of the test charge, making it a property of just the other charges. As a result, voltage is defined as the energy "per unit charge".

Look at the definition of the electric field. Notice that it, to, is defined to be a property of space, not a property of the test charge -- it is force per unit charge. To find the force, you multiply the field intensity at a point by the size of the test charge.

The same with voltage. To find the total work needed to move a test charge from one point to another, you multiply the potential difference by the size of the test charge.

#### kino007

Joined May 24, 2023
7
The force felt by a charge is proportional to the product of both charges. If the test charge is twice as large, force experienced by it due to other charges will be twice as great. However, voltage is defined so that it is independent of the test charge, making it a property of just the other charges. As a result, voltage is defined as the energy "per unit charge".

Look at the definition of the electric field. Notice that it, to, is defined to be a property of space, not a property of the test charge -- it is force per unit charge. To find the force, you multiply the field intensity at a point by the size of the test charge.

The same with voltage. To find the total work needed to move a test charge from one point to another, you multiply the potential difference by the size of the test charge.
Thank you very much for your relpy WBahn. I think i get it now, I'll still have to reread your answer a few times to let it stick but it makes sense. However, when you said: "voltage is defined so that it is independent of the test charge, making it a property of just the other charges" and " Look at the definition of the electric field. Notice that it, to, is defined to be a property of space, not a property of the test charge". Are you implying that the voltage is a property of space? or the "other charges"?

Thanks again for taking the time to answer my question.

#### WBahn

Joined Mar 31, 2012
29,164
Thank you very much for your relpy WBahn. I think i get it now, I'll still have to reread your answer a few times to let it stick but it makes sense. However, when you said: "voltage is defined so that it is independent of the test charge, making it a property of just the other charges" and " Look at the definition of the electric field. Notice that it, to, is defined to be a property of space, not a property of the test charge". Are you implying that the voltage is a property of space? or the "other charges"?

Thanks again for taking the time to answer my question.
A "property of space", particularly in this context, simply means that if you tell me a point in space, I can tell you the value of that property at that point.

This says nothing about what that property depends on. Electric fields (if they are conservative) and the voltage fields that result are properties of space. If the electric fields are not conservative, then this isn't the case and things like voltage become path-dependent, meaning that it's not enough for you to tell me where the point of interest is, but you also need to tell me the path taken to get to that point.

So, given a distribution of charges, I can evaluate the cumulative effect of that distribution and arrive at a vector that describes it for every point in space. I can then use that vector field to define a scalar field that gives the voltage (relative to some arbitrary reference point, which in physics is generally defined to be zero at infinity) at every point in space.

#### kino007

Joined May 24, 2023
7
If the electric fields are not conservative, then this isn't the case and things like voltage become path-dependent, meaning that it's not enough for you to tell me where the point of interest is, but you also need to tell me the path taken to get to that point.
Yea cause the integration done in formulating the voltage in eqn 3 assumed that the path we integrated on was the simplest radial line starting from infinity to a point on the surface of the sphere. I'll have to do some research on non conservative electric fields, that I'm guessing thats when the field is not uniform or symetrical? or is it the case where the field lines simply don't travel in a straight line?

So, given a distribution of charges, I can evaluate the cumulative effect of that distribution and arrive at a vector that describes it for every point in space. I can then use that vector field to define a scalar field that gives the voltage (relative to some arbitrary reference point, which in physics is generally defined to be zero at infinity) at every point in space.
Perfectly stated, understood. Just to clarify, that scalar field that gives the voltage is basically the equipotential lines right?

#### WBahn

Joined Mar 31, 2012
29,164
Yea cause the integration done in formulating the voltage in eqn 3 assumed that the path we integrated on was the simplest radial line starting from infinity to a point on the surface of the sphere. I'll have to do some research on non conservative electric fields, that I'm guessing thats when the field is not uniform or symetrical? or is it the case where the field lines simply don't travel in a straight line?
Because it is path-independent (for conservative electric fields), it doesn't matter which path you perform the integration on, so you try to choose a path that makes the evaluation of the integral as easy as possible.

The definition of a conservative vector field is that it is the gradient of some scalar field. In our case, the electric field is the gradient of the scalar potential (voltage) field. But that probably seems like a rather circular definition to you at this point.

A static electric field is one that is produced by a distribution of charges and it is conservative.

But an electric field that is produced by a changing magnetic field (and is called an "induced electric field" turns out to be nonconservative. It has nothing to do with symmetry or uniformity, but rather how the electric field comes about.

Perfectly stated, understood. Just to clarify, that scalar field that gives the voltage is basically the equipotential lines right?
Equipotential lines are a common way to graphically depict scalar fields of many types, such as lines of equal altitude on a topographic map.

#### kino007

Joined May 24, 2023
7
Because it is path-independent (for conservative electric fields), it doesn't matter which path you perform the integration on, so you try to choose a path that makes the evaluation of the integral as easy as possible.

The definition of a conservative vector field is that it is the gradient of some scalar field. In our case, the electric field is the gradient of the scalar potential (voltage) field. But that probably seems like a rather circular definition to you at this point.

A static electric field is one that is produced by a distribution of charges and it is conservative.

But an electric field that is produced by a changing magnetic field (and is called an "induced electric field" turns out to be nonconservative. It has nothing to do with symmetry or uniformity, but rather how the electric field comes about.

Equipotential lines are a common way to graphically depict scalar fields of many types, such as lines of equal altitude on a topographic map.
I forgot to tell you thanks, I was busy musing the answer you gave. The trouble im having is I always held differentiation in my head as breaking something down to its smallest derivative so that you can find its rate of change or gradient. And, integration has taking the smallest derivative of a thing and adding it up over a specific limit to reconstruct the whole. I know these are crude definitions for integration and differentiation but its how I've been able to "wrap my head around it"... Saying that, thinking of taking the derivative of a scalar field (lower dimensions) and breaking it down (taking the derivative) into a vector field (higher dimensions) does not compute with the current way I see derivatives.

I know the problem lies in the fundamental way I see derivatives, is seeing it as "breaking stuffs down" incorrect?

Again, Thank you for taking the time to share valuable knowledge with me.