PHYSICS-ALEVELS-MCQ HELP-JAN 16 IAL...HELP plss...JAN EXAMS... Watch

MARYAM1234567
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https://pmt.physicsandmathstutor.com...%20A-level.pdf
Why is the answer to Mcq 4 B??
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MARYAM1234567
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also why is MCQ 2 D??
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BobbJo
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(Original post by MARYAM1234567)
https://pmt.physicsandmathstutor.com...%20A-level.pdf
Why is the answer to Mcq 4 B??
2. Electric field follows inverse square law. Charge is negative so electric field strength is negative
so D

4. Consider the force on positive charge,
at midpoint, it is repelled by positive charge and attracted by negative charge

At midpoint, resultant force is

 F = \dfrac{k(Q)q + k(Q)q}{r^2} (add the forces since same direction)

It is not zero

It does not change direction

It is not a maximum: Recall that electric field strength = negative potential gradient
So Force = Charge x Negative Potential Gradient
If you consider derivatives, it is a minimum.

Could not find a standard graph of E or F, so look at this
https://www.desmos.com/calculator/eb5rc2m3lc
Look between values 0 and 3.
It has a minimum at exactly 1.5 (the midpoint)
The graph shows it.
If you differentiate and stuff, you can prove it.

Consider values:
if +Q was 1C

and -Q was 1C

distance between charges was 1 m

at midpoint, force = kq^2/0.5^2 + kq^2/0.5^2 = 71.92 x 10^9N

at 25 cm from positive charge, force = 159.82 x10^9N

at 25 cm from negative charge, i.e 75 cm from positive charge, force = 159.82 x10^9N

hence minimum at midpoint
Last edited by BobbJo; 6 months ago
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MARYAM1234567
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(Original post by BobbJo)
2. Electric field follows inverse square law. Charge is negative so electric field strength is negative
so D

4. Consider the force on positive charge,
at midpoint, it is repelled by positive charge and attracted by negative charge
At midpoint, resultant force is
 F = \dfrac{k(Q)q + k(Q)q}{r^2} (add the forces since same direction)

It is not zero
It does not change direction

It is not a maximum since electric field strength = negative potential gradient
consider derivatives
it is a minimum

Could not find a standard graph, so look at this
https://www.desmos.com/calculator/eb5rc2m3lc
Look between values 0 and 3.
It has a minimum at exactly 1.5
If you differentiate and stuff, you can prove it.

Consider values:
if +Q was 1C
and -Q was 1C
distance between charges was 1 m
at midpoint, force = kq^2/0.5^2 + kq^2/0.5^2 = 71.92N
at 25 cm from positive charge, force = 159.82N
at 25 cm from negative charge, i.e 75 cm from positive charge, force = 159.82N
hence minimum at midpoint
Got it
thanks alot!!
the part where you explained using considering values is genius!....
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Presto
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(Original post by BobbJo)
2. Electric field follows inverse square law. Charge is negative so electric field strength is negative
so D

4. Consider the force on positive charge,
at midpoint, it is repelled by positive charge and attracted by negative charge
At midpoint, resultant force is
 F = \dfrac{k(Q)q + k(Q)q}{r^2} (add the forces since same direction)

It is not zero
It does not change direction

It is not a maximum since electric field strength = negative potential gradient
consider derivatives
it is a minimum

Could not find a standard graph, so look at this
https://www.desmos.com/calculator/eb5rc2m3lc
Look between values 0 and 3.
It has a minimum at exactly 1.5
If you differentiate and stuff, you can prove it.

Consider values:
if +Q was 1C
and -Q was 1C
distance between charges was 1 m
at midpoint, force = kq^2/0.5^2 + kq^2/0.5^2 = 71.92N
at 25 cm from positive charge, force = 159.82N
at 25 cm from negative charge, i.e 75 cm from positive charge, force = 159.82N
hence minimum at midpoint
If you don't mind me asking, can you please tell me how you that 71.92? And those other values? What did you substitute for k and q?
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BobbJo
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(Original post by Presto)
If you don't mind me asking, can you please tell me how you that 71.92? And those other values? What did you substitute for k and q?
I considered the force on a positive charge of 1C
I took values of +Q and -Q
+Q was 1C and -Q was -1C
took the distance between the charges as 1 m
k = 8.99 x 10^9 m/F
(I missed an exponent which is now added)
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Presto
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(Original post by BobbJo)
I considered the force on a positive charge of 1C
I took values of +Q and -Q
+Q was 1C and -Q was -1C
took the distance between the charges as 1 m
k = 8.99 x 10^9 m/F
(I missed an exponent which is now added)
Thank you
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Eimmanuel
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(Original post by BobbJo)
Could not find a standard graph, so look at this
https://www.desmos.com/calculator/eb5rc2m3lc
Look between values 0 and 3.
It has a minimum at exactly 1.5
If you differentiate and stuff, you can prove it.

Consider values:
if +Q was 1C
and -Q was 1C
distance between charges was 1 m
at midpoint, force = kq^2/0.5^2 + kq^2/0.5^2 = 71.92 x 10^9N
at 25 cm from positive charge, force = 159.82 x10^9N
at 25 cm from negative charge, i.e 75 cm from positive charge, force = 159.82 x10^9N
hence minimum at midpoint
I am ok with your explanation but I find that it is an overkilled.
A superposition of 2 graphs can explain why the other 3 options are wrong.

You say the following
If you differentiate and stuff ...
IMO this is confusing. And you did not say what graphs are you showing with desmos.

It may be good that you learn to explain things coherently. If you want to use force, then explain using force. Why do you switch gear using electric field and then switch to electric force?

Sometimes i find some of the smart students like to act smart and want to impress the examiners but it turns out to expose their confusion. My advice to some of my students : Be smart, NOT act smart.
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BobbJo
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(Original post by Eimmanuel)
I am ok with your explanation but I find that it is an overkilled.
A superposition of 2 graphs can explain why the other 3 options are wrong.

You say the following


IMO this is confusing. And you did not say what graphs are you showing with desmos.

It may be good that you learn to explain things coherently. If you want to use force, then explain using force. Why do you switch gear using electric field and then switch to electric force?

Sometimes i find some of the smart students like to act smart and want to impress the examiners but it turns out to expose their confusion. My advice to some of my students : Be smart, NOT act smart.
I am extremely sorry that you found it acting smart. I am not smart, and I cannot act smart.

As I said, I could not find a standard graph showing the E-field strength or force. So I illustrated it with the graph of desmos

The force is proportional to electric field strength. I am extremely sorry at this switching of gear. I went back to answer the question.

Note: I only said to consider the derivative if you want to prove it. The graph already shows the minimum, but if you want to prove it you can differentiate. I do understand the superposition of graphs. I gave it in desmos curve.

I am extremely sorry

I am also very sorry that the explanation is incoherent.
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Eimmanuel
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(Original post by BobbJo)
I am extremely sorry that you found it acting smart. I am not smart, and I cannot act smart.

As I said, I could not find a standard graph showing the E-field strength or force. So I illustrated it with the graph of desmos

The force is proportional to electric field strength. I am extremely sorry at this switching of gear. I went back to answer the question.

Note: I only said to consider the derivative if you want to prove it. The graph already shows the minimum, but if you want to prove it you can differentiate. I do understand the superposition of graphs. I gave it in desmos curve.

I am extremely sorry

I am also very sorry that the explanation is incoherent.
There is really nothing for you to be sorry.

When I quote the particular statement:
If you differentiate and stuff, you can prove it.
I find it confusing is because what do you really meant by “If you differentiate and stuff”.

I am also not sure what you really meant by a standard graph of an electric field. Look at the following link:
https://en.wikipedia.org/wiki/File:E...e_changing.gif
I believe the animated graph illustrates the resultant electric field between 2 charges beautifully. Perhaps, what frustrates the viewers is that it is a bit difficult to see the graph properly when it is changing.

On the hand, I think you are smart (IMO). Ensure that you understand what you read and don’t follow thing blindly. Learn to discern!
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Eimmanuel
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(Original post by BobbJo)
I am extremely sorry that you found it acting smart. I am not smart, and I cannot act smart.

As I said, I could not find a standard graph showing the E-field strength or force. So I illustrated it with the graph of desmos

The force is proportional to electric field strength. I am extremely sorry at this switching of gear. I went back to answer the question.

Note: I only said to consider the derivative if you want to prove it. The graph already shows the minimum, but if you want to prove it you can differentiate. I do understand the superposition of graphs. I gave it in desmos curve.

I am extremely sorry

I am also very sorry that the explanation is incoherent.
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