Torque problem

Below is a figure of a long, nonconducting, massless rod of length L , pivoted at its center and balanced with a block of weight W at a distance x from the left end. At the left and right ends of the rod are attached small conducting spheres with a positive charges q and 2q , respectively. A distance h directly beneath each of these spheres is a fixed sphere with a positive charge Q . I am going to show how to find the distance x when the rod is horizontal and balanced and the value h should have so that the rod exerts no vertical force on the bearing when the rod is horizontal and balanced.

In order for the Rod to be horizontal and balanced, it must be in equilibrium. Because 2 electrostatic forces and 1 weight force cause torque about the axis, the sum of these three torques must add to equal 0 which is notated as \sum\Upsilon = 0

The equation for the electromagnetic force, F_1 , is \frac{qQ}{4\pi \epsilon_0 h^2} , and the equation for the electromagnetic force, F_2 is \frac{2qQ}{4\pi\epsilon_0 h^2} , where q and Q represent respective charges, h is the distance between spheres, and \epsilon_0 is the permittivity constant which is a measure of the resistance that is encountered when forming an electric field in a medium.

Now after adding up the torques and setting them equal to zero I obtain this equation: (\frac{qQ}{4\pi\epsilon_0 h^2})(\frac{L}{2}) + W(x-\frac{L}{2}) - (\frac{2qQ}{4\pi\epsilon_0 h^2})(\frac{L}{2}) = 0 . After solving for Wx and factoring out \frac{L}{2} out of the right side of the equation, I obtain the equation as follows: Wx = \frac{L}{2}(\frac{2qQ}{4\pi\epsilon_0 h^2} + W - \frac{qQ}{4\pi\epsilon_0 h^2}) . Now after combining like terms and dividing each side by the weight variable W , we get the final solution for x which is as follows: x= (\frac{L}{2})(\frac{qQ}{4\pi\epsilon_0 h^2W} + 1)

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