Electric field strength Watch

Shaanv
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Hi guys.

Im looking at question 2c part ii.

So in the previous parts of the question i have worked out the charge is negatively charged and i worked out the magnitude of the charge.

Im fine with the calculations however i am confused as to why they have only used the magnitude of the charge as opposed to the charge in c part ii.

So my question is should the answer be the negative of what they have got or am i making a mistake somewhere.

[link]http://pmt.physicsandmathstutor.com/...tential QP.pdf[/link]
[link]http://pmt.physicsandmathstutor.com/...tential MS.pdf[/link]

Thanks in advance.

Edit: i apologise i forgot an i. Im looking at 2c part iii. Thats probz pretty important😂.
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phys981
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Possibly because it just says strength? As in magnitude?
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Shaanv
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(Original post by phys981)
Possibly because it just says strength? As in magnitude?
But E the electric field strength is defined as the force per unit positive charged on a test mass placed in the field at that point. A positive and a negative charge attract, so the force would be negative so doesn’t that mean the electric field strength would be negative?

The questions annoy me sometimes😤
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phys981
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Isn't part (ii) refering to a postive charge anyway?
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Shaanv
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(Original post by phys981)
Isn't part (ii) refering to a postive charge anyway?
As i understand it the charge producing the field we are talking about is a negative charge.

The question does mention a positive charge being moved in the field however the electric field strength isnt affected as it is the force per unit charge acting on a small positive test charge in the field.

Its possible that im making a mess of this, im gonna have a break then cone back to it and see if i have any new insights. Ill keep u posted👊🏾
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phys981
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I think part (c) (ii) is referring to the same +60nC charge as part (c) (i) and that's why they haven't used a negative with the charge.
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Shaanv
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(Original post by phys981)
I think part (c) (ii) is referring to the same +60nC charge as part (c) (i) and that's why they haven't used a negative with the charge.
Mark scheme says no marks if Q=60nC
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Eimmanuel
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(Original post by Shaanv)
Hi guys.

Im looking at question 2c part ii.

So in the previous parts of the question i have worked out the charge is negatively charged and i worked out the magnitude of the charge.

Im fine with the calculations however i am confused as to why they have only used the magnitude of the charge as opposed to the charge in c part ii.

So my question is should the answer be the negative of what they have got or am i making a mistake somewhere.

[link]http://pmt.physicsandmathstutor.com/...ntial%20QP.pdf[/link]
[link]http://pmt.physicsandmathstutor.com/...ntial%20MS.pdf[/link]

Thanks in advance.

I am not sure about what do you mean that “you are fine with the calculations however you are confused as to why they have only used the magnitude of the charge as opposed to the charge in c part ii”

I suspected that you don’t really understand what the MS is doing. The MS has actually used the negative sign of the charge implicitly in one of the calculation and explicitly in another calculation.


The use of the negative sign of the charge implicitly (IMO) is

at r = 0.20 m V = −1250V and at r = 0.50m V = −500V
so pd ΔV = −500 − (−1250) = 750 (V)
work done ΔW (= QΔV) = 60 × 10 −9 × 750
= 4.5(0) × 10 −5 (J) (45 μJ)



The use of the negative sign of the charge explicitly (IMO) is

 \Delta V = -\dfrac{27.8 \times 10^{-9} }{4 \pi \epsilon_0 } \times (\dfrac{1}{0.5} - \dfrac{1}{0.2}) = \dfrac{27.8 \times 10^{-9} }{4 \pi \epsilon_0 } \times (\dfrac{1}{0.2} - \dfrac{1}{0.5})
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phys981
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(Original post by Shaanv)
Mark scheme says no marks if Q=60nC
Sorry, I've been looking at part (ii) - that's what I thought you were looking at too :/

(Original post by Shaanv)
Im looking at question 2c part ii.



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Shaanv
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(Original post by Eimmanuel)
I am not sure about what do you mean that “you are fine with the calculations however you are confused as to why they have only used the magnitude of the charge as opposed to the charge in c part ii”

I suspected that you don’t really understand what the MS is doing. The MS has actually used the negative sign of the charge implicitly in one of the calculation and explicitly in another calculation.


The use of the negative sign of the charge implicitly (IMO) is

at r = 0.20 m V = −1250V and at r = 0.50m V = −500V
so pd ΔV = −500 − (−1250) = 750 (V)
work done ΔW (= QΔV) = 60 × 10 −9 × 750
= 4.5(0) × 10 −5 (J) (45 μJ)



The use of the negative sign of the charge explicitly (IMO) is

 \Delta V = -\dfrac{27.8 \times 10^{-9} }{4 \pi \epsilon_0 } \times (\dfrac{1}{0.5} - \dfrac{1}{0.2}) = \dfrac{27.8 \times 10^{-9} }{4 \pi \epsilon_0 } \times (\dfrac{1}{0.2} - \dfrac{1}{0.5})
Im on about the very last part of question 2 where it asks us to calculate the electric field strength.

I thought it would be negative as in the first part of the question the potential is negative as the charge is negative.

I did what the markscheme did except i used -27.8nC as opposed to 27.8nC

Edit: just realised i forgot an i in my intial post. Im looking at 2c part iii.
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Shaanv
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It was my fault i missed an i in my first post sorry about that.🍻
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Eimmanuel
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(Original post by Shaanv)
Im on about the very last part of question 2 where it asks us to calculate the electric field strength.

I thought it would be negative as in the first part of the question the potential is negative as the charge is negative.

I did what the markscheme did except i used -27.8nC as opposed to 27.8nC

Edit: just realised i forgot an i in my intial post. Im looking at 2c part iii.

It seems that the question is asking for magnitude and this often occurred in A level physics question. Even if you include the negative sign, the markers will ignore it.
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Shaanv
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(Original post by Eimmanuel)
It seems that the question is asking for magnitude and this often occurred in A level physics question. Even if you include the negative sign, the markers will ignore it.
So to conclude, IRL the field strength would be negative, however the mark scheme is only interested in magnitude.

Thanks
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Eimmanuel
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(Original post by Shaanv)
So to conclude, IRL the field strength would be negative, however the mark scheme is only interested in magnitude.

Thanks
Not sure why I just get an alert on this.

I am not sure what you mean by “IRL”. You should not generalize the result as it depends on the situation. To me, the negative sign does not mean anything to me unless a coordinate system is stated.

A lot of standard A level physics tends to write the following for gravitational force:

 F = -\dfrac{GMm}{r^2}

and say that the gravitational force is attractive because of the negative sign. In general, this is not correct. You need to understand the context of the writing instead of just generalize the result.
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