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Two point charges suspended in a horizontal electric field?
Two 1.27 g spheres are suspended by 34.4 cm long light strings
A uniform electric field is applied in the x direction. If the spheres have charges of +4.21×10-8 C and -4.21×10-8 C, determine the magnitude of the electric field intensity that enables the spheres to be in equilibrium at q = 20.5o
Is q the angle between them? Or the angle with equilibrium?
I’ll assume that the angle is from vertical and that the charges don’t interact with one another.
Horizontal force = electrostatic force – horizontal component of tension
= qE – T sin (theta) = 0Vertical force = vertical compenent of tension – weight
= T cos (theta) – mg = 0Two equations and two unknowns–T and E.
Solve the second equation for the tension:
T = mg / cos(theta)And plug that into the first
qE – mg tan (theta) = 0Solve for the electric field:
E = m g tan (theta) / qThey give you the mass, angle, and charge. You know g. Plugnchug.
EDIT: If you don’t want to make the assumption that the interaction between the charges is small (and maybe you shouldn’t), then they will attract each other with a force
F = k q^2 / (r sin(theta)^2)So the horizontal equation becomes
qE – T sin (theta) – kq^2 / (r sin(theta)^2) = 0
That just gives you another term to your answer when you solve for the electric field. You’ll have to plug in the numbers to see if that second term is significant.
E = m g tan (theta) / q + kq / (r sin(theta)^2)
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