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TheSoftAttack
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Homework Statement
Two small spheres, each of mass 4.00g, are suspended by light strings 15cm in length. A uniform electric field is applied in the x direction. The spheres have charges equal to -7*10^-8 C and +7*10^-8 C. Determine the electric field that enables the spheres to be in equilibrium at an angle of [tex]\theta[/tex] = 13[tex]^{o}[/tex].
Homework Equations
F = [tex]\frac{k_{e}Q_{1}Q_{2}}{r^{2}}[/tex]
F = mgsin[tex]\theta[/tex]
The Attempt at a Solution
So, for this problem I thought it would be a good idea to calculate the force of the two charged spheres attracting each other. This is given by the equation F = [tex]\frac{k_{e}Q_{1}Q_{2}}{r^{2}}[/tex]. We already know what Q1 and Q2 are, and k is a constant. To calculate r, we just take the area between the two strings, and divide it into two right triangles. That way, we can use the Pythagorean theorem to get that (r/2) = (The length of the string)(sin[tex]\theta[/tex]) = 0.15sin13. Now we have:
F = (8.99*10^9)(-7*10^-8)(7*10^-8)/((2*0.15sin13)^2) = -0.00967N
I also calculated the horizontal component of the gravitational force, which is given by F=mgsin[tex]\theta[/tex]. This gives us (4*10^-3)(9.8)(sin13) = 0.00882N
From the perspective of the negatively charged sphere, both of these forces are acting in the same direction (pulling the sphere to the right), so I added the absolute value of these two quantities to get (0.00967 + 0.00882) = 0.01849N
Because the horizontal forces on the charged sphere need to add up to zero for the sphere to be in equilibrium, there needs to also be a force of 0.01849N pushing the sphere to the left. This force will be caused by the uniform electric field.
To get the magnitude of this electric field, we use the equation F = qE.
E = F/q = (-0.01849N)/(-7*10^-8) = 264,151 N/C.
This is the answer I got, but it's not correct. If anyone knows what I am doing incorrectly / what I actually need to be doing, I'd appreciate any help.