Electric field on two charged spheres

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Homework Help Overview

The problem involves two charged spheres with equal but opposite charges suspended in a horizontal electric field. The spheres are at rest, forming an angle due to the forces acting on them, and the goal is to determine the magnitude of the electric field.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the forces acting on the spheres, including the electric force and the attractive force between the oppositely charged spheres. There are questions about whether the attraction force needs to be accounted for and how it affects the equilibrium position of the spheres.

Discussion Status

Some participants have provided insights into the forces involved and the need to consider both the electric and attractive forces. There is an ongoing exploration of the relationships between these forces and their contributions to the angle formed by the strings.

Contextual Notes

Participants note the convention of charge placement and the implications for the direction of forces. There is also a mention of the need for clarity on which charge is positioned where, as this affects the analysis of forces.

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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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