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Coulomb force

Can someone help me on this last one i got no idea..

I have calculated the distance r^2 as:

a^2-(a^2/4)

Now I dont know what to do next..

Am i right in saying that because we have a net charge of zero the electric fields must be equal?

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(edited 10 years ago)

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Reply 1

Stonebridge

Yes the resultant electric field (force) must be zero.

As Q must be negative if the charges at B and C are positive, all you have to do is equate the size of the force on the charge at A due to the charge Q to the size of the resultant force due to the charges at B and C. Use Coulomb's Law.

Reply 2

a10

Original post by Stonebridge

Thanks, can you help me on the equating part because i can't seem to get any of the answers :s-smilie:

First part:

Unparseable latex formula:

F = k.Q.q \div \fraction 3/4 a^2

Second part:

F = k.q^2 \div a^2

are these correct?

Edit: Also does the electric field act in any point in space inside the triangle (as in it can't be zero at just one point or can it)?

(edited 10 years ago)

Reply 3

Stonebridge

Original post by a10

Thanks, can you help me on the equating part because i can't seem to get any of the answers :s-smilie:

First part:

Unparseable latex formula:

F = k.Q.q \div \fraction 3/4 a^2

Second part:

F = k.q^2 \div a^2

are these correct?

Edit: Also does the electric field act in any point in space inside the triangle (as in it can't be zero at just one point or can it)?

Yes but you need the resultant of two forces due to the charges at B and C in the direction from A to Q. So you need the cos of some angles in that second equation.

Reply 4

a10

Original post by Stonebridge

Yes but you need the resultant of two forces due to the charges at B and C in the direction from A to Q. So you need the cos of some angles in that second equation.

okay so just to clarify: the second part should be multiplied by cos(60) and the first part by cos(30)? Kinda confused on this part..

i don't understand why we are considering the force at B to C as this force is not in the same direction as the force from A to Q? Or are we just using the fact that inside the whole triangle the resultant force will be zero regardless of the vector direction on individual point charges?

Edit, this is what i tried to solve:

F = 4kQq\div 3a^2 \times \sqrt 3 \div 2

F = kq^2\div a^2 \times 1 \div 2

then made Q the subject and ended up with:

Q = (-) q ( 3\div 4 \sqrt 3 )

(and added the minus sign since Q must be negative)

seems i have got the reciprocal of the answers though..

(edited 10 years ago)

Reply 5

Stonebridge

What I'm saying is this
For there to be zero force on the charge at the top, the attractive force due to Q (F_Q) must be balanced by the two repulsive forces (F) due to the positive charges q at B and C. Q must be negative.
So put the magnitude of the coulomb force on A due to Q, equal to the sum of the two components of the forces F in the upwards direction. So that's 2Fcos 30