# Homework Help: Two equal positive charges are held fixed and separated by d

1. Aug 18, 2016

### Hamdi Allam

1. The problem statement, all variables and given/known data
Two equal positive charges are held fixed and separated by distance D. Find the electric field vector along their perpendicular bisector. Then find the position relative to their center where the field is a maximum

There was another thread with this question but the suggestions/answer is unclear

3. The attempt at a solution

From symmetry, should E along the i^ direction be 0 along the perpendicular bisector ?

and since the charges is the same, the E vector should be E = 2Kq/(y^2) j^. Am i correct for the first part of this question? Y is the distance along the perpendicular bisector relative to the center of the two charges.

I am stuck at the second part of this question

2. Aug 18, 2016

### Staff: Mentor

You need to add the electric fields associated with each of the two charges vectorially. What is the y component of each of these fields (in terms of D and y)?

3. Aug 18, 2016

### Hamdi Allam

Hi,

Is this regarding the second portion of the question or the first?

4. Aug 18, 2016

### Staff: Mentor

Both.

5. Aug 18, 2016

### Hamdi Allam

Would it be kq/y^2? I am not sure how D would be incorporated in the y component of the field

6. Aug 18, 2016

### Staff: Mentor

What would be the magnitude of the force exerted by the charge at (-D/2,0) on a positive test charge q* located at (0,y)? What would be the x and y components of this force?

7. Aug 18, 2016

### Hamdi Allam

would the magnitude be: K*q*q/((d^2)/4 + y^2)
x component: K*q*q/((d^2/4)
y component: K*q*q/(y^2)

8. Aug 18, 2016

### Staff: Mentor

The magnitude is correct, but the components are incorrect. You need to reconsider the trigonometry.

9. Aug 18, 2016

### Hamdi Allam

ohhhhhh, silly mistake

the x component would be: k*q*q*D/(2*((d^2)/4 + y^2)^3/2))
the y component would be: k*q*q*y/((d^2)/4 + y^2)^3/2)

Is this correct?

10. Aug 18, 2016

yes