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Resultant electric field strength question

Hi, can someone give me some pointers on how to solve this?
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@titfortat @Kyx @DarkEnergy (If any of you guys don't want me to tag you in questions then just let me know haha)
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(edited 7 years ago)

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Original post by jessyjellytot14
Hi, can someone give me some pointers on how to solve this?
@DarkEnergy

Sorry not sure how to do this
Original post by jessyjellytot14
Hi, can someone give me some pointers on how to solve this?
ImageUploadedByStudent Room1480882819.074144.jpg



@titfortat @Kyx @DarkEnergy (If any of you guys don't want me to tag you in questions then just let me know haha)
Posted from TSR Mobile


Use the electric field strength of a point charge formula
E=kqr2 E = \frac{kq}{r^2}

Place the q1=+8.0nC at the origin.

Assume it is a distance d from the origin that the vector sum of the electric field strength is zero.

k(8.0nC)d2k(2.0nC)(60d)2=0 \frac{k*(8.0 nC)}{d^2} - \frac{k*(2.0 nC)}{(60-d)^2} = 0

Solve for d. There are two solutions and you need to know which to reject.
Original post by Eimmanuel
Use the electric field strength of a point charge formula
E=kqr2 E = \frac{kq}{r^2}

Place the q1=+8.0nC at the origin.

Assume it is a distance d from the origin that the vector sum of the electric field strength is zero.

k(8.0nC)d2k(2.0nC)(60d)2=0 \frac{k*(8.0 nC)}{d^2} - \frac{k*(2.0 nC)}{(60-d)^2} = 0

Solve for d. There are two solutions and you need to know which to reject.

Yupp this is what I did ^^ Just remember that the 60 written is supposed to be 60 x 10^-3 (as its in mm) and k being 1/(4pi*epsilon0)

Original post by jessyjellytot14
Hi, can someone give me some pointers on how to solve this?
ImageUploadedByStudent Room1480882819.074144.jpg

@titfortat @Kyx @DarkEnergy (If any of you guys don't want me to tag you in questions then just let me know haha)
Posted from TSR Mobile


Original post by DarkEnergy
Sorry not sure how to do this
(edited 7 years ago)
Original post by titfortat
Yupp this is what I did ^^ Just remember that the 60 written is supposed to be 60 x 10^-3 (as its in mm)


It does not matter. If you do the math correctly, the units would take care by itself.
Original post by titfortat
Yupp this is what I did ^^ Just remember that the 60 written is supposed to be 60 x 10^-3 (as its in mm) and k being 1/(4pi*epsilon0)

I did this method but it gave me a quite big quadratic. I know you can just use a graphical calculator or the quadratic formula but it seemed like quite a lot of work for just one mark, thought there would've been an easier way :x
Original post by DarkEnergy
I did this method but it gave me a quite big quadratic. I know you can just use a graphical calculator or the quadratic formula but it seemed like quite a lot of work for just one mark, thought there would've been an easier way :x


Meh thats multiple choice qs for you! So much time for only one mark
Original post by DarkEnergy
I did this method but it gave me a quite big quadratic. I know you can just use a graphical calculator or the quadratic formula but it seemed like quite a lot of work for just one mark, thought there would've been an easier way :x


k(8.0nC)d2k(2.0nC)(60d)2=0 \frac{k*(8.0 nC)}{d^2} - \frac{k*(2.0 nC)}{(60-d)^2} = 0

This can be simplified to

8.0d22.0(60d)2=0 \frac{8.0}{d^2} - \frac{2.0}{(60-d)^2} = 0

OR

4d2=1(60d)2 \frac{4}{d^2} = \frac{1}{(60-d)^2}

You plot 4x2 \frac{4}{x^2} and 1(60x)2 \frac{1}{(60-x)^2} respectively.
And find the points of intersection
Original post by DarkEnergy
I did this method but it gave me a quite big quadratic. I know you can just use a graphical calculator or the quadratic formula but it seemed like quite a lot of work for just one mark, thought there would've been an easier way :x


If you want an easier way, it is to do elimination. The point where the electric field strength is zero has to be nearer to the point charge with less charge. This would help to reduce the choices from 4 to 2.

Since the equation has been set up, just plug in one of the remaining choices to see if it is the solution. No solving of quadratic equation!
Original post by jessyjellytot14
Hi, can someone give me some pointers on how to solve this?
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Suppose E=0 when the distance from the +2 nC is r1r_1 and the distance from the +8 nC is r2r_2. Call this point P. We know that:

E1=2kr12,E2=8kr22E_1=\frac{2k}{r_1^2}, E_2=\frac{8k}{r_2^2}

At P, we must have E1=E2E_1=E_2 so that:

2kr12=8kr22\frac{2k}{r_1^2} = \frac{8k}{r_2^2}

We also know that r1+r2=60 mmr_1+r_2=60 \text{ mm}

Can you finish this off?
Reply 10
Original post by jessyjellytot14
Hi, can someone give me some pointers on how to solve this?
ImageUploadedByStudent Room1480882819.074144.jpg



@titfortat @Kyx @DarkEnergy (If any of you guys don't want me to tag you in questions then just let me know haha)
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One can just use ratios. If one knows that Charge 2 is 4 times larger than charge 1, then the distance from charge 2 would be 4 times greater than from charge 1.
Reply 11
Original post by Kyx
One can just use ratios. If one knows that Charge 2 is 4 times larger than charge 1, then the distance from charge 2 would be 4 times greater than from charge 1.


So then one has the following simultaneous equations:

A + B = 60
A - 4B = 0

Subtract equation 2 from equation 1 to leave us with:

3B = 60
B = 20

Substituting into equation 1, one gets:

A = 40

The answer is 40mm
Original post by Kyx
One can just use ratios. If one knows that Charge 2 is 4 times larger than charge 1, then the distance from charge 2 would be 4 times greater than from charge 1.


You certainly can't do this, since it doesn't take the inverse square law into account.
Original post by Kyx
So then one has the following simultaneous equations:

A + B = 60
A - 4B = 0

Subtract equation 2 from equation 1 to leave us with:

3B = 60
B = 20

Substituting into equation 1, one gets:

A = 40

The answer is 40mm


A=40A=40 and B=20B=20 aren't solutions to your original equations, since we then have

A4B=404(20)=4080=40A-4B = 40-4(20) = 40-80 = -40

You made a mistake when subtracting, but regardless of that, you won't get the right answer since it turns out that the distances are in the ratio 2:1, not 4:1. See my post above.
Reply 14
Original post by atsruser
You certainly can't do this, since it doesn't take the inverse square law into account.


Original post by atsruser
A=40A=40 and B=20B=20 aren't solutions to your original equations, since we then have

A4B=404(20)=4080=40A-4B = 40-4(20) = 40-80 = -40

You made a mistake when subtracting, but regardless of that, you won't get the right answer since it turns out that the distances are in the ratio 2:1, not 4:1. See my post above.


Thx
Original post by Eimmanuel
Use the electric field strength of a point charge formula
E=kqr2 E = \frac{kq}{r^2}

Place the q1=+8.0nC at the origin.

Assume it is a distance d from the origin that the vector sum of the electric field strength is zero.

k(8.0nC)d2k(2.0nC)(60d)2=0 \frac{k*(8.0 nC)}{d^2} - \frac{k*(2.0 nC)}{(60-d)^2} = 0

Solve for d. There are two solutions and you need to know which to reject.


LOL my teacher said you'll never have to use quadratics in A-level physics :dong:
I'll try and solve this in a bit though, thanks!
Original post by jessyjellytot14
LOL my teacher said you'll never have to use quadratics in A-level physics :dong:
I'll try and solve this in a bit though, thanks!


Have you looked at posting no. 9?
Original post by Eimmanuel
Have you looked at posting no. 9?


Yes - How would you set up the elimination method?
Original post by jessyjellytot14
Yes - How would you set up the elimination method?


Let E1 E_1 be the magnitude of the electric field strength due to the +8.0 nC point charge and E2 E_2 be the magnitude of the electric field strength due to the +2.0 nC point charge.

The point where E1=E2 E_1 = E_2 should be nearer to the +2.0 nC point charge which means A and B could be eliminated.
Reply 19
Post 10 solved this.

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