Question

Two tiny spheres of mass = 8.70 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 between the strings equal to 50 in the following figure. (note that q1 is positive and it is the left once)

http://session.masteringphysics.com/problemAsset/1331112/7/21-82.jpg

What is the magnitude E of the field?

Answer 1

Here is a free-body diagram of the forces on q1The sphere has the force of gravity, the tension of the string, the force of attraction due to the other charge q2, and the force due to the induced electric field.
The vector sum of these forces is zero\[\sum\vec F_1=\vec Fg+\vec T+\vec F_2+\vec F_E=0\]Breaking the vectors into x and y components gives a system of equations
for the x-direction:\[\sum F_{1x}=F_{Ex}+F_{2x}+T_x=q_1E+\frac1{4\pi\epsilon_0}\frac{q_1q_2}{r^2}+T_x\]a little trig and we can see from the figure that the distance between charge 1 and 2 is\[r=2L\sin(\frac\theta2)\]so our formula becomes\[\sum F_{1x}=qE-\frac1{16\pi\epsilon_0}\frac{q^2}{L^2\sin^2(\frac\theta2)}-T\sin(\frac\theta2)=0\]and for the y-direction:\[\sum F_{1y}=T_{1y}+F_g=T\cos(\frac\theta2)-mg=0\]we can eliminate the unknown tension of the string by solving the second equation for T and subbing it into the first, which gives\[\sum F_{1x}=qE-\frac1{16\pi\epsilon_0}\frac{q^2}{L^2\sin^2(\frac\theta2)}-mg\tan(\frac\theta2)=0\]Which we can use to solve for E.