Answer :
The tension in the cable is F = πEaV²r³/2d²
The work. done. by the cable. is W = πEaV²r²/4d
[tex]\mathbf{U_i-U_f=\frac{\varepsilon_0\pi r^2 V^2}{2d}-\frac{\varepsilon_0\pi r^2 V^2}{4d}}\mathbf{\Rightarrow U_i-U_f=\frac{\varepsilon_0\pi r^2 V^2}{4d}}[/tex]
The equation can be used to depict the electric field created between two circular plates of a capacitor if they are separated by a constant distance and supported by a cable.
E = V/d
E' = E/2
F = qv/2d
F = Eav2r2/2d2.
Assuming that dz has shifted the upper plate higher, we can express the tension's work as follows:
dw = T*dz
dz = F * dw
The cable's work is W = Eav2r2/4d, or W = Eav2r2/2z2*dz.
Before the top plate is raised, the electrical energy capacitor has the following amount of energy stored in it:
Ui = 1/2 Cv²
Ui = EaR2V2D
Yes, the work carried out by the cable is equivalent to the change in electrical energy that has been stored. The variation in energy previously stored
Ui = Eaπr²v²/4d
[tex]\mathbf{U_i-U_f=\frac{\varepsilon_0\pi r^2 V^2}{2d}-\frac{\varepsilon_0\pi r^2 V^2}{4d}}\mathbf{\Rightarrow U_i-U_f=\frac{\varepsilon_0\pi r^2 V^2}{4d}}[/tex]
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