AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

Original post by TheSmartOne

Hi guys, I'm struggling with Question 13 from the MC section of June 2010, can someone help me out? Sorry if someone else has already asked about it, I just don't have the time to go through the thread!

The diagram shows two charges, +4x10^-6C and -16x10^-6C, 120mm apart. What is the distance from the +4x10^-6C charge to the point between the two charges where the resultant electric potential is zero?

A) 24mm
B) 40mm
C) 80mm
D) 96mm

I would do it as a ratio
so with the charges youve got a 4:16 ratio which is equal to a 1:4 ratio
1+4 = 5, 120mm divided by 5 gives 24 mm

This method will only work wwith linear problems though i.e potentials dont try it with field strengths or force as theyre quadratics

Reply 2724

Leonacatherine

Original post by TheSmartOne

Hi guys, I'm struggling with Question 13 from the MC section of June 2010, can someone help me out? Sorry if someone else has already asked about it, I just don't have the time to go through the thread!

The diagram shows two charges, +4x10^-6C and -16x10^-6C, 120mm apart. What is the distance from the +4x10^-6C charge to the point between the two charges where the resultant electric potential is zero?

A) 24mm
B) 40mm
C) 80mm
D) 96mm

Just as a side not if you are ever unsure you can use a bit of common sense to make an educated guess. If they were equal charges the resultant would be equal to zero at the midpoint, but because the -16x10^-6 is much larger than 4x10^-6 in magnitude you know the point where the resultant is equal to zero is going to be closer to the 4 one. This immediately eliminates C and D as theyre closer to the 16 one :smile:

Reply 2725

OP

Original post by Leonacatherine

I would do it as a ratio
so with the charges youve got a 4:16 ratio which is equal to a 1:4 ratio
1+4 = 5, 120mm divided by 5 gives 24 mm

This method will only work wwith linear problems though i.e potentials dont try it with field strengths or force as theyre quadratics

You can do it with quadratics. Just square root the ratio as the negative solution can be rejected.

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Reply 2726

saad97

Can anyone help me with this question please?

Reply 2727

Leonacatherine

Original post by saad97

Can anyone help me with this question please?

F = qq/ 4 x pi x E0 x r^2
Because the magnitude of q is the same qq = q squared
4x pi is just a number and so has no units
so Q^2/ E0 x r^2
so d :smile:

x

Reply 2728

OP

Original post by saad97

Can anyone help me with this question please?

Coulomb's law states that, for two charges in an electric field:

[br]F = \dfrac{Qq}{4 \pi \epsilon_0 d^2}[br]

In this case the charges are equal in magnitude, so:

[br]F = \dfrac{Q^2}{4 \pi \epsilon_0 d^2}[br]

As

4 \pi

is a constant, it can be ignored when looking at units as it doesn't have any.

As such, removing this from the equation, this means:

[br]\dfrac{Q^2}{\epsilon_0 d^2}[br]

Must have the same units as F.
Does that help?

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Reply 2729

saad97

Original post by Leonacatherine

F = qq/ 4 x pi x E0 x r^2
Because the magnitude of q is the same qq = q squared
4x pi is just a number and so has no units
so Q^2/ E0 x r^2
so d :smile:

x

Original post by CD223

Coulomb's law states that, for two charges in an electric field:

[br]F = \dfrac{Qq}{4 \pi \epsilon_0 d^2}[br]

In this case the charges are equal in magnitude, so:

[br]F = \dfrac{Q^2}{4 \pi \epsilon_0 d^2}[br]

As

4 \pi

is a constant, it can be ignored when looking at units as it doesn't have any.

As such, removing this from the equation, this means:

[br]\dfrac{Q^2}{\epsilon_0 d^2}[br]

Must have the same units as F.
Does that help?

Posted from TSR Mobile

Thank you!
I have another question from the same paper that I need help with please?

Does anyone know if we need to know much about the conservation of kinetic energy, and if there have been any questions on momentum to do with kinetic energy?

Reply 2732

Ilovemaths96

13

defintion of one tesla: magnetic field strength required to exert a force of 1N on a wite of length 1m carrying a current of 1A? Do you guys agree?

Reply 2733

OP

Original post by Ilovemaths96

defintion of one tesla: magnetic field strength required to exert a force of 1N on a wite of length 1m carrying a current of 1A? Do you guys agree?

The tesla can be defined as the force per unit length per unit current that a magnetic field exerts on a conductor in a magnetic field.

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Reply 2734

noseypo

1

Original post by DeiJaVu

Does anyone know if we need to know much about the conservation of kinetic energy, and if there have been any questions on momentum to do with kinetic energy?

I dont think theres much they could ask about that is there?

Reply 2735

OP

Original post by DeiJaVu

Does anyone know if we need to know much about the conservation of kinetic energy, and if there have been any questions on momentum to do with kinetic energy?

There was a MC question (Jan 2013 Q2 I think? Not sure). It was to do with expressing the momentum of a daughter nucleus in terms of kinetic energy.

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Reply 2736

DeiJaVu

Original post by CD223

There was a MC question (Jan 2013 Q2 I think? Not sure). It was to do with expressing the momentum of a daughter nucleus in terms of kinetic energy.

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Okey thankyou, I'll go check that out.

Reply 2737

OP

Original post by saad97

Thank you!
I have another question from the same paper that I need help with please?

Does this help?

ImageUploadedByStudent Room1433615639.017147.jpg

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Reply 2738

Paul Dirac

Original post by DeiJaVu

Does anyone know if we need to know much about the conservation of kinetic energy, and if there have been any questions on momentum to do with kinetic energy?

Kinetic energy is conserved in elastic collisions, and all collisions (including elastic) conserve momentum provided no external resultant force acts on the system. I think that's all.

Reply 2739

Paul Dirac