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

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Why is the answer to Mcq 4 B??

Why is the answer to Mcq 4 B??

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#3

(Original post by

https://pmt.physicsandmathstutor.com...%20A-level.pdf

Why is the answer to Mcq 4 B??

**MARYAM1234567**)https://pmt.physicsandmathstutor.com...%20A-level.pdf

Why is the answer to Mcq 4 B??

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

(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; 1 year ago

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(Original post by

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

(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

**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

(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

thanks alot!!

the part where you explained using considering values is genius!....

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#5

**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

(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

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#6

(Original post by

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?

**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 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)

Last edited by BobbJo; 1 year ago

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#7

(Original post by

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)

**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)

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#8

(Original post by

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

**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

A superposition of 2 graphs can explain why the other 3 options are wrong.

You say the following

If you differentiate and stuff ...

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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#9

(Original post by

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.

**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.

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.

Last edited by BobbJo; 1 year ago

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#10

(Original post by

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.

**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.

When I quote the particular statement:

If you differentiate and stuff, you can prove it.

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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#11

**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.

https://en.wikipedia.org/wiki/File:E...e_changing.gif

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