Question

A metallic circular plate with radius r is fixed to a tabletop. An identical circular plate supported from above by a cable is fixed in place a distance d above the first plate. Assume that dd is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V.


A)What is the tension in the cable? Neglect the weight of the plate.

Express your answer in terms of the variables d, r, V, and constants ϵ0, π.


B)The upper plate is slowly raised to a new height 2d. Determine the work done by the cable by integrating ∫(from d to 2d) F(z)dz, where F(z) is the cable tension when the plates are separated by a distance z.

Express your answer in terms of the variables d, r, V, and constants ϵ0, π.


C)Compute the energy stored in the electric field before the top plate was raised.

Express your answer in terms of the variables d, r, V, and constants ϵ0, π.

D)Compute the energy stored in the electric field after the top plate was raised.
Express your answer in terms of the variables d, r, V, and constants ϵ0, π.

E)Is the work done by the cable equal to the change in the stored electrical energy? If not, why not?
a)The work done in separating the plates is equal to energy change in the plates.
b)The work done in separating the plates is equal to the magnitude of the energy change in the plates. This does not mean that the work done is equal to the change in the energy stored in the plates. The work done on the plates is positive but the plates lose energy. The plates are connected to the battery, so the potential difference across them remains constant as they are separated. Therefore charge is forced off of the plates through the battery, which does work on the battery.

Answer 1

Answer:

the tension in the cable is $\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}$

the work done by the cable is $\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}$

Explanation:

A)

If we have two circular plate supported by a cable at a fixed distance, then the electric field formed between the two plate of the capacitor can be represented by the equation.

$\mathbf{E = \frac{voltage \ \ V}{distance \ \ d}}$

However; the net electric field i.e the sum of the electric filed produced is represented as:

$\mathbf{E' = \frac{E}{2}} \\ \\ \mathbf{E' = \frac{V}{2d}}$

So, if we assume that the lower plate and the upper plate possess the charge +q and -q respectively. Then, the tension of the cable which is the same as Force F can be written as:

$\mathbf{F = q* E'}$

$\mathbf{F = \frac{q*v}{2d}}$ -----    equation (1)

Also ; we know that

$\mathbf{C = \frac{q}{v}= \frac{E_oA}{d}}$

$\mathbf{\frac{q}{v}= \frac{E_o \pi r^2}{d}} \ \ \ \ \ \mathbf{since \ A = \pi r^2}$

$\mathbf{{q}= \frac{\pi E_o {v} r^2}{d}}$    -----   equation (2)

Replacing equation 3 into equation (2); we have:

$\mathbf{F = \frac{\pi E_o vr^2}{d}* \frac{v}{2d}}$

$\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}$

Therefore,  the tension in the cable is $\mathbf{F = \frac{\pi E_o v^2r^2}{2d^2}}$

B)

Assume that the upper plate is displaced by dz in an upward direction ; Then we can express the workdone by the tension as :

$\mathbf{dW = T *dz} \\ \\ \mathbf{dW = F*dz} \\ \\ \mathbf{dW = \frac{\pi E_o v^2r^2}{2z^2}dz }$

The net workdone to raise the plate from separation d to 2d is:

$\mathbf{W = \int\limits^{2d}_{2zd} {dw} = \frac{\pi E_ov^2r^2}{2} \int\limits^{2d}_d \frac{dz}{z^2} }$

$\mathbf{W= \frac{\pi E_ov^2r^2}{2} [-\frac{1}{z}]^{2d}_d }$

$\mathbf{W= - \frac{\pi E_ov^2r^2}{2} [\frac{1}{2d}-\frac{1}{d}]}$

$\mathbf{W= - \frac{\pi E_ov^2r^2}{2} [\frac{-1}{2d}]}$

$\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}$

the work done by the cable is $\mathbf{W= \frac{\pi E_ov^2r^2}{4d}}$

C) To calculate the energy stored in the Electrical energy Capacitor before the top plate is raised ; we have:

$\mathbf{U_i = \frac{1}{2}Cv^2} \\ \\ \mathbf{U_i = \frac{1}{2}(\frac{E_oA}{d})v^2} \\ \\ \mathbf{U_i = \frac{1}{2}(\frac{E_o \pi r^2}{d})v^2} \\ \\ \mathbf{U_i = \frac{E_o \pi r^2 v^2}{2d}} }$

D) The energy stored in the plate after the  the top plate was raised is as follows:  

$\mathbf{U_f = \frac{1}{2}C'v^2} \\ \\ \mathbf{U_f = \frac{1}{2}(\frac{E_oA}{2d})v^2} \\ \\ \mathbf{U_f = \frac{1}{2}(\frac{E_o \pi r^2}{2d})v^2} \\ \\ \mathbf{U_f = \frac{E_o \pi r^2 v^2}{4d}} }$

E) Yes,  work done by the cable equal to the change in the stored electrical energy. The Difference in energy stored before and after the top plate is raised:

$\mathbf{U_i-U_f} = \mathbf{\frac{E_o \pi r^2 v^2}{2d}} }} - \mathbf {\frac{E_o \pi r^2 v^2}{4d}} }}$

$\mathbf{U_i-U_f}= \mathbf {\frac{E_o \pi r^2 v^2}{4d}} }}$

Thus;

b)The work done in separating the plates is equal to the magnitude of the energy change in the plates. This does not mean that the work done is equal to the change in the energy stored in the plates.