Electric field on two charged spheres

AI Thread Summary
Two charged spheres, each with a mass of 7.90 mg and a charge of 72.0 nC of opposite signs, are suspended and form an angle of 50.0 degrees when a horizontal electric field is applied. The discussion emphasizes the need to consider both the electric force and the attractive force between the oppositely charged spheres to determine the total force acting on each sphere. It is clarified that the electric field direction is to the left, with the positive charge on the left and the negative charge on the right. The participants highlight the importance of understanding the vector components of the forces involved to solve for the electric field's magnitude. The conversation concludes with a reminder that assistance is provided for understanding rather than completing homework problems.
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Homework Statement


Two tiny spheres of mass m = 7.90 mg carry charges of equal magnitude, 72.0 nC, but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 m. When a horizontal uniform electric field E that is directed to the left is turned on, the spheres hang at rest with the angle θ between the strings equal to 50.0 degrees in the following figure.
What is the magnitude of E?

Homework Equations


qE = F

The Attempt at a Solution


I know that I need the total force acting on a sphere in order to calculate the electric field strength. I just can't seem to solve for that force.
Does the attraction force of the oppositely charged spheres need to be accounted for as well? Or is the angle separation only caused by the electric field force?
Thanks for helping me out here!
 

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Yes, the attraction force needs to be accounted for. What will the extra electric force do? Well, actually, even for myself to answer that question I'd need to know which charge is on the left/right so hopefully you're given that.
 
Mindscrape said:
Yes, the attraction force needs to be accounted for. What will the extra electric force do? Well, actually, even for myself to answer that question I'd need to know which charge is on the left/right so hopefully you're given that.

The charge on the left is positive, and the charge on the right is negative. We know this because the convention for my class is that electric fields come outward from positive charges.

Ok, so since there is an attractive force, then at the equilibrium shown, would it just be ƩF= F_{E} - F_{q} ?
Or: force due to electric field - force due to attraction = the total force causing the angle ?
 
Oh, yeah, duh, the electric field has to be repelling because the coulomb force is attractive in this case, which means positive on left and negative on right.. I was just checking to make sure you were paying attention. <_< >_>

You're getting there. You're missing another important force vector. What is the vector that the electric forces are giving? What else contributes to making the angle?
 
Assume Electric field intenisty due to Charge +Q around it is constant
and it's value is 5N/C and then an other +q0 charge of 2C is place in +Q charge field +q0 experience force that is F=5*2=10N because(F=Eq0)

Q:tell me how much force is required to move +q0 charge against electric filed to 1m?
 
No, you tell me how much force is required. People here help with homework, not do homework. :)
 
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