Electric Forces

[andie]

I need help answering this question:

 

A chage +q is located at the origin, while an identical charge is located on the x-axis at x= +0.50m. A third charge of +2q is located on the x-axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

 

I need some pointers on how to answer the question.

 

PLEASE HELP!

[Klaynos]

Firstly draw the system and work out the field at the origin.

[andie]

yes i do have a diagram for it, but now I need to know how to attempt solving the problem.

[swansont]

Do you have an equation you can apply to the situation?

[chemhelper]

Force = kQq/(r^2) Q=charge at point P and q=charge a distance r from P

 

So the initial force would equal k(Q^2)/.25

 

The problem specifies that the force doubles as a result of its placement. Thus, the force in question becomes 2k(Q^2)/.25 = 8k(Q^2)

 

Set this equal to the same force equation at the beginning

 

8k(Q^2) = kQq/(r^2) little q = 2Q by specification. as you can see, the Q^2 and k drops out and what is left is 8 = 2/(r^2)

 

You'll get two answers for R. Try to figure out which side of the charge placing 2Q will increase the force on the charge as opposed to counteracting the force due to the charge at 0.5 m

[NeonBlack]

http://www.wholikeshomework.com

 

ask them

 

chemhelper are you the owner of that website?

[swansont]

The goal here is not to do the homework for people who ask homework questions, but to help them do it for themselves. A service that sells homework solutions is basically facilitating plagiarism.

[Klaynos]

First work out the force at the origin due to the first 2 charges. Using the electrostatic force formula for point charges, Coulomb's law:

 

http://en.wikipedia.org/wiki/Coulomb's_law#Vector_form

 

For more than one charge you add them together, so the first 2 charges force at the origin added together timesed by 2 should be = to the 3 charges added together..

 

Which is pretty much what chemhelper said.