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the tension in the cable is the work done by the cable is 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.However; the net electric field i.e the sum of the electric filed produced is represented as: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...

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matthewdabber7

15.10.2021

Physics

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:

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

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.

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