Electric field on two charged spheres

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SUMMARY

The discussion focuses on calculating the magnitude of the electric field (E) affecting two charged spheres, each with a charge of 72.0 nC and a mass of 7.90 mg. The spheres are suspended and form an angle of 50.0 degrees due to the influence of a horizontal electric field. Participants emphasize the necessity of accounting for both the electric force and the attractive force between the oppositely charged spheres to determine the total force acting on each sphere.

PREREQUISITES
  • Understanding of Coulomb's Law and electric forces
  • Knowledge of vector addition in physics
  • Familiarity with the concept of electric fields and their directionality
  • Basic principles of equilibrium in physics
NEXT STEPS
  • Calculate the electric field strength using the formula E = F/q
  • Explore the concept of force vectors in equilibrium scenarios
  • Study the effects of electric fields on charged particles in different configurations
  • Review the principles of electric force interactions between multiple charges
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Students studying electromagnetism, physics educators, and anyone interested in understanding the dynamics of charged particles in electric fields.

Nivlac2425
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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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