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 dd 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.1) 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, pi.F=2)
The upper plate is slowly raised to a new height 2d. Determine the work done by the cable by integrating ∫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,pi.
W=
3) 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, pi.
U=
4) 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, pi.
U=
5)
is the work done by the cable equal to the change in the stored electrical energy? If not, why not?



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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