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AQA A2, Electric field strengths and equal points

There is a couple of questions i have seen in exams which the books i have don't teach me how to do, and i cannot for the life of me figure out what to do,

1: A +15x10^-6 charge q1 is a distance of 20mm from q2 which is 10x10^-6 Coulombs

•

How do you calculate the resultant elecrtic field strength and

•

why is there a point along the 20mm line where the resulting electric field strength is zero,

•

and how is this point calculated

Trying to do the last question resulted in me getting a quadratic and for a 1 mark answer, im sure there's a simpler way of doing it

2: If there are two point masses attracting eachother, and they're around 9000km apart, will there be a point along the line where the resultant field strengths are zero, eg when an object placed between them is equally attracted to both masses so no net movement occurs

3: A small object O carrying a charge +Q is placed at a distance d from a metal plate that has an
equal and opposite charge. The object is acted on by an electrostatic force F. How can the force on this object be expressed mathematically, as my books dont tell me what to do if a point charge is placed near a plate

4: two charges, +4 µC and –16 µC, 120 mm apart. What is the distance from
the +4 µC charge to the point between the two charges where the resultant electric potential is
zero?

Thanks in advance

(edited 12 years ago)

Reply 1

Stonebridge

13

First take a look at the answers to the question in this recent thread.
http://www.thestudentroom.co.uk/showthread.php?t=2201683

There will be a point somewhere between two masses, or two positive or two negative charges where the resultant field strength will be zero. This is because electric and gravitational field strength are vectors, and there will be a point where the two forces are equal in magnitude but opposite in direction.
You find the point by putting the magnitude of the force at that point due to the one charge (or mass) equal to the magnitude of the other. If the distance between the charges is d and you say the point is a distance r from the one charge then it's a distance d-r from the other.
For electric potential
You get an equation for r in the form
X/d = Y/(d-r)
were X, Y and d will be values given in the question for charge, the constant and the separation d.
The simplest way to solve it is to turn it upside down. It certainly isn't quadratic.
Edit
Please see my next post for electric field strength.

(edited 2 years ago)

Reply 2

mar junior

OP

Thanks for the reply, So in your equation X/d = Y/(r-d)

X = charge
Y = Constant
d = Seperation
R = total distance?

Even if this is true it will not work if the charges are different, how would you use the equation if the values for d and x are different?

From what i see electric filed strength:

KQ/R^2. So for two fields and finding the point where field strengths are equal:

-(KQ/R^2) = KQ/(R-X)^2

Where R is the total distance between both charges and X is the only unknown. and trying to rearrange this results in a quadratic

(edited 12 years ago)

Reply 3

Stonebridge

13

Original post by mar junior

Thanks for the reply, So in your equation X/d = Y/(r-d)

X = charge
Y = Constant
d = Seperation
R = total distance?

Even if this is true it will not work if the charges are different, how would you use the equation if the values for d and x are different?

From what i see electric filed strength:

KQ/R^2. So for two fields and finding the point where field strengths are equal:

-(KQ/R^2) = KQ/(R-X)^2

Where R is the total distance between both charges and X is the only unknown. and trying to rearrange this results in a quadratic

Yes. There was a typo in my original post, the bracket should have been (d-r)
The equation is of the form X/r = Y/(d-r)
This is actually for the point where the electric potential is zero, not the force (field strength). Potential goes with 1/r and field strength with 1/r²
Apologies for not checking my answer before pressing the send button.

The equation for potential (the one I gave) is not quadratic. The one for field strength of course is quadratic as it has a r squared term in it.
You have the correct equation there in your post for the field strength.

However...
The quick way of doing these (without the quadratic) is to realise that because the constant cancels on both sides you get

\frac{Q_1}{R_1^2} = \frac{Q_2}{R_2^2}

R1 is the distance of Q1 from the zero point, and R2 the distance of Q2
Which gives the ratio of the distances as

\frac{R_1^2}{R_2^2} = \frac{Q_1}{Q_2}

\frac{R_1}{R_2} = \sqrt{\frac{Q_1}{Q_2}}

An easy example would be two charges of 16C and 9C separated by 14mm
The 14mm distance between them would divided in the ratio 4 to 3, (square roots of the two charges) giving the zero point 8mm from the 16C and 6mm from the 9C

Reply 4

mar junior

OP

Thank you so much, your help is invaluable, the ratio trick you devised helped, and it works with all the calculations i do,

Do you know how long i have been trying figure out these questions for? i tried every rearrangement and formula imaginable and you do it in one go, How did you know to use ratios in this situation?
Also will this technique work for potentials?

(edited 12 years ago)

Reply 5

Stonebridge

13

Original post by mar junior

Thank you so much, your help is invaluable, the ratio trick you devised helped, and it works with all the calculations i do,

Do you know how long i have been trying figure out these questions for? i tried every rearrangement and formula imaginable and you do it in one go, How did you know to use ratios in this situation?
Also will this technique work for potentials?

It's actually a well known trick.
It's even easier for the potential zero point because the ratio of the distances is just the ratio of the charges. However, for the zero point to be somewhere between the two charges, one must be plus and the other minus.

Reply 6

mar junior

OP

is it? well i never would have known, thanks again

Reply 7

username3416616

12

Original post by Stonebridge

Yes. There was a typo in my original post, the bracket should have been (d-r)
The equation is of the form X/r = Y/(d-r)
This is actually for the point where the electric potential is zero, not the force (field strength). Potential goes with 1/r and field strength with 1/r²
Apologies for not checking my answer before pressing the send button.

The equation for potential (the one I gave) is not quadratic. The one for field strength of course is quadratic as it has a r squared term in it.
You have the correct equation there in your post for the field strength.

However...
The quick way of doing these (without the quadratic) is to realise that because the constant cancels on both sides you get

\frac{Q_1}{R_1^2} = \frac{Q_2}{R_2^2}

R1 is the distance of Q1 from the zero point, and R2 the distance of Q2
Which gives the ratio of the distances as

\frac{R_1^2}{R_2^2} = \frac{Q_1}{Q_2}

\frac{R_1}{R_2} = \sqrt{\frac{Q_1}{Q_2}}

An easy example would be two charges of 16C and 9C separated by 14mm
The 14mm distance between them would divided in the ratio 4 to 3, (square roots of the two charges) giving the zero point 8mm from the 16C and 6mm from the 9C

i believe the root trick is now patched?

Reply 8

Stonebridge

13

Original post by itslitbro

i believe the root trick is now patched?

Yes that was 9 years ago. There was a typo in my first post but I corrected it.
It's worth remembering this shortcut, especially for MCQs where you get just one mark and are on limited time.

AQA A2, Electric field strengths and equal points

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