# Very tricky physics question.

## Homework Statement

An equilateral triangle has a point charge +q at each of the three vertices (A, B, C). Another point charge Q is placed at D, the midpoint of the side BC. What is the charge Q if the total electric force on the charge at A due to the charges at B, C, and D is zero? (Use any variable or symbol stated above as necessary.)

Image: http://www.webassign.net/grr/p16-24.gif

Q= _______

kq/r^2

## The Attempt at a Solution

The total net force on A is zero.

Let's say the length of AB and AC are d so that the distance of B and C to A are d. Now, we have to find the distance between Q and A. Well, Q is placed in the midpoint of BC, so that we can bisect the triangle by drawing a line through A and Q. That makes a 30-60-90 triangle in which the hypotenuse is d. Using trigonometry, we know that QB and QC are d/2, making QA (d*sqrt(3))/2.

Coulomb's law:

[ k(+q)(+q)/d² ] + [ k(+q)(+q)/d² ] + [ k(Q)(+q)/((d*sqrt(3))/2)² ]
To make the math easier, take out the common factor k(+q) and solve ((d*sqrt(3))/2)
k(+q) [ (+q)/d² + (+q)/d² +(Q)/(3d²/4) ]

So, +2q/d² + (Q)/(3d²/4) = 0
+2q/d² = -(Q)/(3d²/4)
Multiply 3d² and divide 4 to the other side
+2q(3d²)/4d² = -Q
d² cancels out, multiply both sides by -1
Q = -(3/2)q

Webassign says it's wrong. Can you think of any way I can rearrange it? Or is it done entirely wrong?

gneill
Mentor
[ k(+q)(+q)/d² ] + [ k(+q)(+q)/d² ] + [ k(Q)(+q)/((d*sqrt(3))/2)² ]

You haven't accounted for the fact that the forces from the charges at B and C on the charge at A will have x and y components. The forces will add as vectors.

## Homework Statement

An equilateral triangle has a point charge +q at each of the three vertices (A, B, C). Another point charge Q is placed at D, the midpoint of the side BC. What is the charge Q if the total electric force on the charge at A due to the charges at B, C, and D is zero? (Use any variable or symbol stated above as necessary.)

Image: http://www.webassign.net/grr/p16-24.gif

Q= _______

kq/r^2

## The Attempt at a Solution

The total net force on A is zero.

Let's say the length of AB and AC are d so that the distance of B and C to A are d. Now, we have to find the distance between Q and A. Well, Q is placed in the midpoint of BC, so that we can bisect the triangle by drawing a line through A and Q. That makes a 30-60-90 triangle in which the hypotenuse is d. Using trigonometry, we know that QB and QC are d/2, making QA (d*sqrt(3))/2.

Coulomb's law:

[ k(+q)(+q)/d² ] + [ k(+q)(+q)/d² ] + [ k(Q)(+q)/((d*sqrt(3))/2)² ]
To make the math easier, take out the common factor k(+q) and solve ((d*sqrt(3))/2)
k(+q) [ (+q)/d² + (+q)/d² +(Q)/(3d²/4) ]

So, +2q/d² + (Q)/(3d²/4) = 0
+2q/d² = -(Q)/(3d²/4)
Multiply 3d² and divide 4 to the other side
+2q(3d²)/4d² = -Q
d² cancels out, multiply both sides by -1
Q = -(3/2)q

Webassign says it's wrong. Can you think of any way I can rearrange it? Or is it done entirely wrong?

should take only y components while calculating - x components will cancel out themselves
dont think of forces as scalars

supratim1
Gold Member
just take y components of charges at B and C. note that Q has just y component. equate them, get Q = -q.

gneill
Mentor
just take y components of charges at B and C. note that Q has just y component. equate them, get Q = -q.

supratim1
Gold Member
2*(kq^2)cos60 = - kQq
=> q = -Q
=> Q = -q

gneill
Mentor
2*(kq^2)cos60 = - kQq
=> q = -Q
=> Q = -q

1. The cosine gives the x-component, not the y-component.
2. The distances to the charge at A are not the same.

supratim1
Gold Member
sorry, a mistake by me. its cos30, not cos 60.
and yes, i forgot about the different distances due to my haste, i apologize.

But the cosine gives y component, when the side BA and CA are extended.

so, now,

(2(kq^2)cos 30)/(a^2) = - 4kQq/(3a^2)

solve.

Q = - 3(sqrt 3)q/4