Electric/gravitation potential

Original post by Georgeleeds

The graphs of the attractive electric field and gravitational field

The field strength of gravitation is the negative of the gradient, so will be negative (-dv/dr)

The field strength of the attractive electric field will be positive, (dv/dr)

Why do they have different signs if they are both attractive??

Thanks

Posted from TSR Mobile

It's really just the result of the fact that gravitation is always attractive and historically didn't need to distinguish between attraction and repulsion.
Electric fields have positive and negative charges and so we can distinguish the two types of force using the "likes repel and opposites attract", idea.
Because mass doesn't have a minus associated with it, masses are just always considered positive, then you have "likes attract" in gravitation. In order to make it the same mathematically to electrical forces, you would have to have one mass plus and the other minus.
So the answer is that it's just the way the maths works out because gravitation and electrical forces, though similar in most respects, differ in this one. It isn't anything to get too concerned over so long as you are aware of this.

Reply 2

Georgeleeds

OP

So what does a negative field mean? Compared to a positive field?

E.g -9.81 N/kg
9.81 N/kg

Posted from TSR Mobile

Reply 3

Original post by Georgeleeds

So what does a negative field mean? Compared to a positive field?

E.g -9.81 N/kg
9.81 N/kg

Posted from TSR Mobile

It usually just represents direction.
The one with the minus will be in the opposite direction to the one with the plus.
But you need to provide the context to that example to get a definite answer.

Reply 4

Original post by Stonebridge

It usually just represents direction.
The one with the minus will be in the opposite direction to the one with the plus.
But you need to provide the context to that example to get a definite answer.

Hi Stonebridge, I saw your posts for solutions for physics and the way you answer the questions is exactly how I wish my physics teacher had taught me. Unfortunately, I didn't know how to send you a direct message so I'll post it here because I'm really struggling with A2 multiple choice. I think its because I haven't been taught the basic principles of the topics such as gravitational and electric fields. There's only so much I can learn from a book :frown:

Anyway, I saw your solution to a question about resultant electric field strength between 2 point charges and can't thank you enough!

Just to remind you, here's your answer: "The quick way of doing these (without the quadratic) is to realise that because the constant cancels on both sides you get

R1 is the distance of Q1 from the zero point, and R2 the distance of Q2
Which gives the ratio of the distances as

So the distance between the charges is divided in the ratio of the square roots of the charges.
8 / 2 = 4 and root 4 is 2 so the distance between them is in the ratio 2 to 1
This means it must be 40mm to 20mm giving 2 to one and 60mm in total.

Multiple choice for these will usually give a simple ratio for the charges.
Given the time allowed you need a quick way of doing it without having to solve quadratics.
So
Find the sq root of the ratio of the charges and divide the distance up in that proportion."

I was just wondering that if the two points charges were not positively charged (so if one was positive and the other negative) would you tackle the question in the same way?

Thank you so much! Also sorry for posting on here, I didn't know how else to contact you.

:smile:

Reply 5

You won't get a question like this with the charges one positive and one negative because in that situation there is no zero-force point between the charges.
On the other hand you may get a question about a zero potential point. In this case there is one between the charges when one is + and the other -.
the answer is very similar. The only difference is that potential depends on just the inverse of distance, and not distance squared. So you do the same and divide the distance up but in the straight inverse ratio of the charges not the inverse square.
Remember, this is the quick method for multi choice questions where time is short. If you are doing this in a longer question for 2 or 3 marks, say, you would need to show your calculation and where the formula comes from as I did in my original post.

(edited 9 years ago)

Reply 6

Original post by Stonebridge

You won't get a question like this with the charges one positive and one negative because in that situation there is no zero-force point between the charges.
On the other hand you may get a question about a zero potential point. In this case there is one between the charges when one is + and the other -.
the answer is very similar. The only difference is that potential depends on just the inverse of distance, and not distance squared. So you do the same and divide the distance up but in the straight inverse ratio of the charges not the inverse square.
Remember, this is the quick method for multi choice questions where time is short. If you are doing this in a longer question for 2 or 3 marks, say, you would need to show your calculation and where the formula comes from as I did in my original post.

Thank you so much!! Yes this is for my multiple choice paper. I haven't really had a teacher for the whole year so I've been teaching myself the whole time. Which is why I find it very difficult to make certain assumptions and short cuts with questions like this :frown:

Thank you so much for your help! So can I apply this same method to other topics in the course of is it simply electric fields?

(edited 9 years ago)

Reply 7

Original post by YOOZERNAYME

Thank you so much!! Yes this is for my multiple choice paper. I haven't really had a teacher for the whole year so I've been teaching myself the whole time. Which is why I find it very difficult to make certain assumptions and short cuts with questions like this :frown:

Thank you so much for your help! So can I apply this same method to other topics in the course of is it simply electric fields?

It also applies to the zero gravitational force point between two masses.
The force depends on the inverse square of the distance in the same way as it does for charge. Just do the same for the two masses. You sometimes get a question on this asking about such a point between the earth and the moon. However, these questions are not usually multi choice and would need the full solution. The masses are not usually simple ratios like the charges. Even so it's worth keeping the idea in mind. The principle is the same.

Reply 8

Original post by Stonebridge

It also applies to the zero gravitational force point between two masses.
The force depends on the inverse square of the distance in the same way as it does for charge. Just do the same for the two masses. You sometimes get a question on this asking about such a point between the earth and the moon. However, these questions are not usually multi choice and would need the full solution. The masses are not usually simple ratios like the charges. Even so it's worth keeping the idea in mind. The principle is the same.

Thank you! You've cleared that up a lot for me!

Reply 9

Georgeleeds

OP

Original post by Stonebridge

It usually just represents direction.
The one with the minus will be in the opposite direction to the one with the plus.
But you need to provide the context to that example to get a definite answer.

But if both of them are attractive (gravitational and attractive electric field) then why would the fields be in the opposite direction, wouldn't the field lines for both direct towards the centre?

Thanks

Posted from TSR Mobile

Reply 10

Original post by Georgeleeds

But if both of them are attractive (gravitational and attractive electric field) then why would the fields be in the opposite direction, wouldn't the field lines for both direct towards the centre?

Thanks

Posted from TSR Mobile

I answered this in post #2
* See below.

Your previous question

So what does a negative field mean? Compared to a positive field?

E.g -9.81 N/kg
9.81 N/kg

(and my answer)

referred to two values
9.8N/kg and -9.8N/kg
these are both gravitational.

Not, as you are now asking above, gravitational and electric.

* Think of a graph with a positive charge at the origin.
Place a second positive test charge along the x axis in the plus x direction.
The force on this test charge is repulsive and in the plus x direction.
So the force is positive.
Put a negative charge at the origin and the force is attractive. It acts on the test charge in the negative x direction. The force is negative.
The sign of the force is not so much about attraction and repulsion, but more about direction.

If you do the same for gravity and masses it doesn't work.
#Put a mass at the origin and a test mass somewhere along the x axis in the positive x direction.
The force is (always) attractive but acts towards the origin in the negative x direction. So this force is negative using the same convention.
As I said in post #2.
This is a consequence of the fact that gravitational and electric forces, though similar in many ways, differ in this respect.

(edited 9 years ago)

Reply 11