# Electric Field Due To Dipole

Math Electricity Dipole

6 replies to this topic

### #1 Nishan

Nishan

Advanced Member

• Members
• 33 posts

Posted 03 July 2019 - 08:24 PM

Why component od E perpendicular to r is:
Etheta= 1/r*dv/dtheta

### #2 exchemist

exchemist

Creating

• Members
• 2671 posts

Posted 04 July 2019 - 10:50 AM

Why component od E perpendicular to r is:
Etheta= 1/r*dv/dtheta

Because this is resolving the field, which is a vector quantity,  into component vectors, in terms of polar coordinates. It is just an alternative to doing the same thing in terms of the usual Cartesian (x and y axis) components of distance from an origin, but using radius r and angle θ instead.

Have you not encountered polar coordinates before? If not, you could try reading this to see if it helps you: https://en.wikipedia...ordinate_system

If you draw a dipole on a piece of paper and then draw electric field lines from it, you will see they are elliptical and not circular.  If you now place the origin of a coordinate system at the centre of the dipole, you can draw a series of radial lines for r and circumferential lines for θ. You will very easily see that, in general, the direction of the field line at any point, will not be parallel to either the radial lines or the circumferential lines of your polar coordinate system.

That means the field vector at that point has both radial (r ) and circumferential (θ) components.  That is what is going on.

If you do not know these mathematical concepts, I advise you to learn them before you read more of this book. You will find very frustrating if you try to learn mathematical physics before you have equipped yourself with the mathematical toolkit required. I have the impression this book may not be very good at explaining concepts visually, before it goes into the mathematical treatment.

Edited by exchemist, 04 July 2019 - 10:58 AM.

### #3 Nishan

Nishan

Advanced Member

• Members
• 33 posts

Posted 05 July 2019 - 10:54 PM

Because this is resolving the field.....

I thought I replied but forgot to press post reply. Sorry for that.

### #4 exchemist

exchemist

Creating

• Members
• 2671 posts

Posted 06 July 2019 - 02:33 AM

I thought I replied but forgot to press post reply. Sorry for that.

I'm travelling for a few days so away from my books. Suggest checking field strength versus potential. I suspect this may explain the 1/r factor that bothers you, but I'd prefer to confirm rather than rely my fading memory of electrostatics.
I will try to respond more fully when I can.

### #5 Nishan

Nishan

Advanced Member

• Members
• 33 posts

Posted 08 July 2019 - 02:29 AM

I'm travelling for a few days so away from my books. Suggest checking field strength versus potential. I suspect this may explain the 1/r factor that bothers you, but I'd prefer to confirm rather than rely my fading memory of electrostatics.
I will try to respond more fully when I can.

Whenever you are free

### #6 exchemist

exchemist

Creating

• Members
• 2671 posts

Posted 11 July 2019 - 05:04 AM

I thought I replied but forgot to press post reply. Sorry for that.

OK, I'm back at my laptop now.

I've had a look and I see what you mean. However, I think there is a step here your text book has skipped.

The electric field strength is the rate of change of potential with respect to distance. The distance perpendicular to r is not given by θ, but by rθ, i.e. the portion of the circumference of a circle at radius r (for example it is 2πr when θ = 2π).

So the rate of change of V with distance perpendicular to r is dV/d(rθ), which is the same as dV/r dθ, since r is constant. So that is where your factor of 1/r comes from, I think.

However, as a mere chemist, on things like this I'm always open to correction by a real physicist, if we have any on the forum. (I think sanctus and OceanBreeze may be)

### #7 Nishan

Nishan

Advanced Member

• Members
• 33 posts

Posted 11 July 2019 - 08:03 AM

OK, I'm back at my laptop now.

I've had a look and I see what you mean. However, I think there is a step here your text book has skipped.

The electric field strength is the rate of change of potential with respect to distance. The distance perpendicular to r is not given by θ, but by rθ, i.e. the portion of the circumference of a circle at radius r (for example it is 2πr when θ = 2π).

So the rate of change of V with distance perpendicular to r is dV/d(rθ), which is the same as dV/r dθ, since r is constant. So that is where your factor of 1/r comes from, I think.

However, as a mere chemist, on things like this I'm always open to correction by a real physicist, if we have any on the forum. (I think sanctus and OceanBreeze may be)

Yes thats great man thank you