Example 3- Charge Pendulums
In this example, let’s consider two tiny conducting balls with identical mass, m, and identical charge, q, hang from nonconducting threads, let’s say of length L.
If we draw the picture that we have two pendulums, they are in equilibrium, making an angle of θ with the horizontal with equal masses of m and equal charges of q, and let’s denote the distance between the charges, or the masses, as x. Assume that the angle θ is so small that tangent of θ can be approximated to sine θ, which is a common approximation used for small angles, and the question is show that the distance x is equal to q2L where L is the length of the pendulums divided by 2πε0 times m times g, quantity power of one-third.
In order to analyze these type of systems which are in equilibrium, first we identify the forces and then apply the equilibrium conditions. If we recall the equilibrium conditions, the first one was the sum of the forces should add to 0. This conditions simply avoids the system to accelerate. And the second condition was sum of the torques with respect to any axis of rotation should be equal to 0, and this condition basically guarantees no rotation for our system. And of course second condition becomes irrelevant if we are dealing with point masses, which is our case. Therefore we will only apply the first condition.
To be able to do that, first we will identify the forces acting each one of these objects. Of course the first one will be the gravitational force due to the earth’s pull in downward direction. Since we have a string in the system, we will have a tensional force acting along the string pointing from the end of the string towards its center.
Now, if these are the only two forces acting on this object, then we would never end up with the equilibrium because if we visualize a coordinate system such that the mass is located at the center, at the origin I should say, and by resolving the tensional force into its components we will have a vertical component and as well as a horizontal component. Although the vertical component will be in opposite direction with the weight of the object so that they can cancel, the horizontal component will never be balanced to put the system in equilibrium. Therefore, we have to have another force acting on this object to put the system or to put the object into equilibrium, and that force will come basically due to the charge nature of the other pendulum.
Since both of them are likely charged, then this object will be repelled by the other pendulum to the right direction. Of course, similar type of forces will be acting on the other object: its weight in downward direction, tensional force along the string, and also the repulsive Coulomb force exerted by the other pendulum. Let’s denote the Coulomb force as Fc.
All these forces are going to be equal in magnitude-wise because the masses are identical, charges are identical, the identical strings, et cetera. If we isolate one of these masses and apply the equilibrium conditions, then we’re going to have enough expressions to be able to solve or calculate whatever the quantity that we’re searching for.
To do that, let’s consider this mass and introduce a coordinate system such that the mass is located at the origin, an xy coordinate system. Since we are after with sum of the forces, and from the vector addition rule we can add the forces or add the vectors which lie along the same axis, we introduce this xy coordinate system and then resolve all the force vectors into their components relative to this coordinate system. If you do that, we see only tensional force will have two components, x and y, because Fc, the Coulomb force, will be lying along x-axis only and gravitational force mg will be lying along y-axis only.
Now let’s go ahead and apply the equilibrium condition for our problem. Sum of the forces along x direction should add up to 0. This condition will give us that the Coulomb force, which is in positive x direction, plus x component of the tensional force, which is in negative x direction, therefore times –î, should add to 0. Sum of the forces along y should be equal to 0 will give us y component of the tensional force, which is in positive y direction, therefore times ĵ, plus gravitational force mg, and that is in negative y direction, therefore times –ĵ, should add up to 0.
From these equations, we can write down that x component of the tensional force should be equal to Coulomb force and y component of the tensional force should be equal to the gravitational force mg. At this step, we will write down the explicit form of these quantities, starting with the x and y components of the tensional force. To be able to do that we will again take the advantage of the right triangles which form when we separate the tensional force into its components.
If this angle is θ, therefore this angle will also be θ. If we consider this green triangle, right triangle, then relative to that triangle, x component of the tensional force will be the opposite side relative to angle θ and the y component will be the adjacent side relative to angle θ.
Therefore Tx is going to be equal to hypotenuse T times sine of angle θ, and similarly Ty will be equal to to times, since it is the adjacent side, cosine of θ. We can express the Coulomb force by using Coulomb’s law. That’ll be equal to 1 over 4πε0 times the product of the magnitude of the charges. q times q will give us q2 divided by the square of the distance separating these two charges, and that distance is given as x. Therefore we will have x2 in the denominator.
Then T sine θ will be equal to 1 over 4πε0 q2 over x2, and y component of the tension, which is T cosine of θ, will be equal to mg. If we solve the second equation over here for tension, which will be equal to mg divided by cosine of θ and substitute this expression for the tension in the first equation, we will have mg over cosine of θ times sine θ, which will be equal to 1 over 4πε0 q2 over x2.
We will have sine θ over cosine θ on the left-hand side of the equation, and that ratio is tangent θ. Therefore the final expression will be mg times tangent of θ is equal to Coulomb constant 1 over 4πε0 times q2 over x2.
Now we will take the advantage of the small angle approximation, which was given in the problem as for small θ tangent θ can be expressed as approximately equal to sine θ. If we go back to our diagram, this distance will be x over 2, and by considering the big triangle over here, if we express the sine of angle θ, which will be equal to the ratio of opposite side, that is x over 2, to the hypotenuse, and that is L, we can express sine θ as x over 2 over L, or simply x over 2L, and this quantity is approximately equal to tangent θ in small angle approximation.
Therefore, substituting this value for the tangent θ in our expression we will have mg x over 2L is equal to 1 over 4πε0 q2 over x2. From this relationship, if we make cross-multiplication, then we will have mg times 4πε0, x times x2 will give us x3 will be equal to 2q2 times L.
We can divide both sides by 2. Then we will end up 2π mg ε0 x3 on the left-hand side of the equation, and solving for x3 we’re going to end up with q2L over 2π ε0 mg. And if you take the cubic root of both sides, we will end up with the expression q2L over 2π ε0 mg to the power of 1 over 3, which is exactly the same equation that we were set to prove.