- #1

Volcano

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I think no. Do we assume the field unchanged?

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- Thread starter Volcano
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- #1

Volcano

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I think no. Do we assume the field unchanged?

- #2

Nick89

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- #3

Volcano

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There are only two charges, so there must be only one force. What will i add to that force?

- #4

Nick89

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Usually this is done with a fictional test charge with a charge of +1C. (Because this simplifies the force equation).

The force that one charge of

[tex]F = \frac{1}{4 \pi \epsilon_0} \frac{ |q|}{r^2}[/tex]

If you have two charges, [tex]q_1, q_2[/tex] than the force is the (vector) addition of the two forces:

[tex]\vec{F} = \vec{F_1} + \vec{F_2}[/tex]

If you need an example, try this:

Two equal positive point charges [tex]q_1 = q_2 = 0.2 \mu \text{C}[/tex] are located at x = 0, y = 0.3m and x = 0, y = -0.3m, respectively. What are the magnitude and direction of the total (net) electric force that these charges exert on a third point charge [tex]Q = 4.0 \mu \text{C}[/tex] at x = 0.4m, y = 0 ?

Consider the following sketch:

The sketch shows the force (represented in it's x and y components) on Q due to the upper charge [tex]q_1[/tex].

From Coulomb's law the magnitude

[tex]F = \frac{1}{4 \pi \epsilon_0} \frac{|q_1Q|}{r^2} = (9.0 \times 10^9) \frac{(4.0 \times 10^-6)(2.0 \times 10^-6)}{0.5^2} = 0.29 \text{N}[/tex]

(the radius r = 0.5 can be found by using pythagoras theorem)

The components of this force are given by:

[tex](F_{\text{1 on Q}})_x = F_{\text{1 on Q}} \text{cos} \alpha = 0.29 \frac{0.40}{0.50} = 0.23 \text{N}[/tex]

[tex](F_{\text{1 on Q}})_y = -F_{\text{1 on Q}} \text{sin} \alpha = -0.29 \frac{0.30}{0.50} = -0.17 \text{N}[/tex]

The same forces can be calculated for the lower charge [tex]q_2[/tex], however, since the system in question is symmetrical you can see directly that the x-component is identical, while the y-component is the same except in the other direction.

So:

[tex]F_x = 0.23 + 0.23 = 0.46 \text{N}[/tex]

[tex]F_y = -0.17 + 0.17 = 0[/tex]

The total force on Q is in the +x-direction with magnitude 0.46 N.

I hope this helps to answer your question!

- #5

Shooting Star

Homework Helper

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I think no. Do we assume the field unchanged?

It is perfectly correct, and the only way in fact, to use the superposition principle to find the intensity at an arbitrary point given any number of charges or charged bodies, provided you know the intensity due to each charge. Here, since both are spherically symmetric charges, outside the charged bodies their fields are as if each is a point charge. So, to find the intensity at a point, use the vector sum of the individual intensities, as Nick89 has explained.

There are only two charges, so there must be only one force. What will i add to that force?

No, you misunderstood. Add the fields due to each charge to get the resultant field at a point.

The force that one charge ofqC exerts on this 1C charge is ofcourse:

[tex]F = \frac{1}{4 \pi \epsilon_0} \frac{ |q|}{r^2}[/tex]

If you have two charges, [tex]q_1, q_2[/tex] than the force is the (vector) addition of the two forces:

[tex]\vec{F} = \vec{F_1} + \vec{F_2}[/tex]

Giving the absolute sign and then taking the vector sum is not correct. Leave out the absolute sign on q.

- #6

Nick89

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(Note the lack of vector arrow above the F in the first equation)

The direction is usually not easy to see directly when you have multiple charges, that's when you use vector addition to find the direction.

- #7

Volcano

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Thank you, but I mean a different thing. There are only two charges(little sphere). The force is between them. Let me try one more.

http://i26.tinypic.com/2mwbh37.jpg

http://i27.tinypic.com/35d6ikz.jpg

Above pictures shows the field lines for a spherical charge. As already known, lines go away uniform for the upper link, so we can say [tex]E \propto \frac{1}{{r^2 }}[/tex]. But for second link, it is not uniform as other. Now can we can say [tex]E \propto \frac{1}{{r^2 }}[/tex] also?

As shown above the field twisted and no more changing with reverse ratio to the square of their distance.

Another say, as you may notice, I add circles which surrond charges. Let's choice two different points on circle for first figure. Any one has the same magnitude of electric field. Now let's do the same thing for second(below) figure. Take two different point on second circle(below figure). Are those two magnitude of field the same? I think no.

http://i26.tinypic.com/2mwbh37.jpg

http://i27.tinypic.com/35d6ikz.jpg

Above pictures shows the field lines for a spherical charge. As already known, lines go away uniform for the upper link, so we can say [tex]E \propto \frac{1}{{r^2 }}[/tex]. But for second link, it is not uniform as other. Now can we can say [tex]E \propto \frac{1}{{r^2 }}[/tex] also?

As shown above the field twisted and no more changing with reverse ratio to the square of their distance.

Another say, as you may notice, I add circles which surrond charges. Let's choice two different points on circle for first figure. Any one has the same magnitude of electric field. Now let's do the same thing for second(below) figure. Take two different point on second circle(below figure). Are those two magnitude of field the same? I think no.

Last edited:

- #8

Shooting Star

Homework Helper

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This is true for any spherically symmetric charge distributions, which means that, if there is a spherically symmetric charge distribution within a sphere, then the field outside the sphere is as if the whole charge is concentrated as a point charge at the centre.

- #9

Nick89

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Can you feel the electric field there yourself? Ofcourse you can't.

The only reason you know that [tex]E \propto \frac{1}{{r^2 }}[/tex] is because, if you hold a test charge of 1C at a distance

You want to know how this proportionality works when you have 2 charges?

I can't tell you directly, but in the same manner that you can add the forces, you can also add the electric fields at a point. Doing this for multiple points might give you the result you are looking for.

- #10

Volcano

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Exactly this.How can you say that [tex]E \propto \frac{1}{{r^2 }}[/tex] without using a test charge?

Can you feel the electric field there yourself? Ofcourse you can't.

The only reason you know that [tex]E \propto \frac{1}{{r^2 }}[/tex] is because, if you hold a test charge of 1C at a distancernear the charge you have drawn, than the force the test charge feels is proportional to 1/r².

This is exactly too. I know this is very easy question. And coulomb Law help to calculate the force. When I look at field lines between two charges, asked myself that, how these twisted field lines already do the same thing like the linear lines. Than thought, maybe we assume something when calculate. Honestly very hard thing to explain with another language by the way here is the best place to learn. Sorry for my weak language but thank you all for help.You want to know how this proportionality works when you have 2 charges?

I can't tell you directly, but in the same manner that you can add the forces, you can also add the electric fields at a point. Doing this for multiple points might give you the result you are looking for.

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