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

https://pmt.physicsandmathstutor.com/download/Physics/A-level/Past-Papers/Edexcel-IAL/Unit-4/January%202016%20(IAL)%20QP%20-%20Unit%204%20Edexcel%20Physics%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 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

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?

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)

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

If you differentiate and stuff ...

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.

If you differentiate and stuff, you can prove it.

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.