'The diagram shows a point charge +q placed in the electric field of a charge +Q

Hi,

I'm having some difficulty with my homework. It's on electric fields.

'The diagram shows a point charge +q placed in the electric field of a charge +Q
The force experienced by the charge +q at point A iss F. Calculate the force experienced by this charge when it is placed at points B,C,D, and E. I neach case explain your answer.'

What is confusing me is I thought the force experienced anywhere in a field is the same. What's more, I don't understand how it can be calculated as no numbers are given?????

please please help I am so confused with how to do this! i have a feeling it might be more mathematical than physics. (i.e. B would be F as only one space is seperating it from +Q. Is this correct)

Reply 1

Stonebridge

13

The force on the charge q will always be given by Coulomb's Law
F=(1/4πεo)Qq/x²
where x is the distance between the charges.
You can't calculate its actual value as you are not given any numbers, you just need to say what x is in each case (in terms of R and r) and substitute it in the formula.
For example, for A, x=R. For B, x=r. (I assume that the size of the grid is R by r, where r is the shorter side. It's not clear from your diagram. You seem to be saying they are both R, but they are clearly not equal.)
So your answer each time is just the formula for Coulomb's Law with the value of x changed so that it is the distance between the charges in terms of R and r.

Reply 2

sportsking17

OP

Stonebridge

The force on the charge q will always be given by Coulomb's Law
F=(1/4πεo)Qq/x²
where x is the distance between the charges.
You can't calculate its actual value as you are not given any numbers, you just need to say what x is in each case (in terms of R and r) and substitute it in the formula.
For example, for A, x=R. For B, x=r. (I assume that the size of the grid is R by r, where r is the shorter side. It's not clear from your diagram. You seem to be saying they are both R, but they are clearly not equal.)
So your answer each time is just the formula for Coulomb's Law with the value of x changed so that it is the distance between the charges in terms of R and r.

Right, well the question says the force experienced by the charge +q at point A is F, so I wrote them relative to F instead of R, but I'm not sure if this is what you're meant to do. And yes, sorry about the diagram, the picture is actually boxes of equal size, I just used excel with different sized boxes, but the dimensioins are R xR.

I have the following answers

B= F (same distance from +Q as A)
C = 2F (twice the distance from +Q as A)
D= 3F (3 times the distance from +Q compared to A)
E= 4F (4 times the distance from +A compared to A)

Do you think this is correct :s-smilie:

OR Is the answer meant to be Coulomb's formula and the F/2F/3F/4F in place of 'x'????

Reply 3

Stonebridge

13

sportsking17

Right, well the question says the force experienced by the charge +q at point A is F, so I wrote them relative to F instead of R, but I'm not sure if this is what you're meant to do. And yes, sorry about the diagram, the picture is actually boxes of equal size, I just used excel with different sized boxes, but the dimensioins are R xR.

I have the following answers

B= F (same distance from +Q as A)
C = 2F (twice the distance from +Q as A)
D= 3F (3 times the distance from +Q compared to A)
E= 4F (4 times the distance from +A compared to A)

Do you think this is correct :s-smilie:

OR Is the answer meant to be Coulomb's formula and the F/2F/3F/4F in place of 'x'????

It's the right approach.
Yes, at B the force is the same as at A because the distance is the same.
At C the distance is 2R so the force is one quarter the value at A. (=F/4). Remember, this (Coulomb's Law) is an inverse square law, so when you double the distance the force is 2 squared (4) times smaller. F is proportional to 1/R².

Reply 4

sportsking17

OP

Stonebridge

It's the right approach.
Yes, at B the force is the same as at A because the distance is the same.
At C the distance is 2R so the force is one quarter the value at A. (=F/4). Remember, this (Coulomb's Law) is an inverse square law, so when you double the distance the force is 2 squared (4) times smaller. F is proportional to 1/R².

Ah, thank you. :smile:

Reply 5

sportsking17

OP

My teacher has marked this homework and I got E wrong. I thought E = F/16 because of the inverse square law, and r is 4 times A. My teacher says the correct answer is F/8.
Can someone please explain why it is F/8?

Reply 6

Stonebridge

13

Because the distance from the charge to E is along the hypotenuse of a triangle with sides 2R and 2R
The hypotenuse squared is (2R)² + (2R)² = 4R² + 4R² = 8R²
This means the force will be 1/8 the value at A

Reply 7

Searcher1

12

Original post by sportsking17

Right, well the question says the force experienced by the charge +q at point A is F, so I wrote them relative to F instead of R, but I'm not sure if this is what you're meant to do. And yes, sorry about the diagram, the picture is actually boxes of equal size, I just used excel with different sized boxes, but the dimensioins are R xR.

I have the following answers

B= F (same distance from +Q as A)
C = 2F (twice the distance from +Q as A)
D= 3F (3 times the distance from +Q compared to A)
E= 4F (4 times the distance from +A compared to A)

Do you think this is correct :s-smilie:

OR Is the answer meant to be Coulomb's formula and the F/2F/3F/4F in place of 'x'????

So what is the value of D?

Reply 8

Callicious

22

Original post by Muzzy357

1/9 F

Don't necropost old threads- same goes for the other guy (Searcher1)

The thread is 11 years old.

'The diagram shows a point charge +q placed in the electric field of a charge +Q

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