Resultant field strength between two charged particles

I have no idea how to get the answer to the question.

Both particles will have a positive field strength, and the field strength can never equal zero. So I can't see how the sum of two postive numbers can ever equal zero?

Reply 1

a10

19

Original post by jon889

I have no idea how to get the answer to the question.

Both particles will have a positive field strength, and the field strength can never equal zero. So I can't see how the sum of two postive numbers can ever equal zero?

at one point the field strength will equal zero. I think the answer is 40mm

the field strength will be zero when the distance is 20mm (the difference between the fields 8-2 = 6) so it must be the same each part hence 60mm / 6 = 10mm for each part so when its 2 the potential difference is 0 so field is 0 and they ask how far away is it from 8 so 60-20 = 40mm. Is that the answer?

(edited 12 years ago)

Reply 2

AS01

12

Original post by a10

at one point the field strength will equal zero. I think the answer is 40mm

the field strength will be zero when the distance is 20mm (the difference between the fields 8-2 = 6) so it must be the same each part hence 60mm / 6 = 10mm for each part so when its 2 the potential difference is 0 so field is 0 and they ask how far away is it from 8 so 60-20 = 40mm. Is that the answer?

I don't know how to do this at all!!
what do u mean by that?

Reply 3

agostino981

Original post by AS01

I don't know how to do this at all!!
what do u mean by that?

We know that the electric field strength is

E=\frac{q}{4\pi \epsilon_0 r^2}

We wish to calculate the point in between the particles such that the field strength is zero, so we set up the following equation.

\frac{q_1}{4\pi \epsilon_0 x^2}=\frac{q_2}{4\pi \epsilon_0 (60\times 10^{-3}-x)^2}

x mm

from one particle is

(60-x) mm

from another particle.

Substitute

q_1

and

q_2

and solve the quadratic equation, reject the impossible case(s), e.g.

x>60mm

and here you go.

(edited 12 years ago)

Reply 4

jon889

OP

5

Thanks for your answer. But isn't that just getting the distance when the two fields are equal and not neccessaroly zero?

If your finding the resultant I thought you have to add them?

Reply 5

Stonebridge

13

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}}

So the distance between the charges is divided in the ratio of the square roots of the charges.
8 / 2 = 4 and root 4 is 2 so the distance between them is in the ratio 2 to 1
This means it must be 40mm to 20mm giving 2 to one and 60mm in total.

Multiple choice for these will usually give a simple ratio for the charges.
Given the time allowed you need a quick way of doing it without having to solve quadratics.
So
Find the sq root of the ratio of the charges and divide the distance up in that proportion.

Reply 6

mac073

1

Original post by Stonebridge

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}}

So the distance between the charges is divided in the ratio of the square roots of the charges.
8 / 2 = 4 and root 4 is 2 so the distance between them is in the ratio 2 to 1
This means it must be 40mm to 20mm giving 2 to one and 60mm in total.
Multiple choice for these will usually give a simple ratio for the charges.
Given the time allowed you need a quick way of doing it without having to solve quadratics.
So
Find the sq root of the ratio of the charges and divide the distance up in that proportion.

Wow, Thank you!

Resultant field strength between two charged particles

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