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 d is much smaller than r. The two plates are attached by wires to a battery that supplies voltage V. Required:
a. What is the tension in the cable?
b. Compute the energy stored in the electric field after the top plate was raised.

A) F = V²E_o•πr²/2d²

B) U = E_o•Aπr²V²/2d

Explanation:

A) Since we have two circular plates, the formula for the electric field is expressed as;

E = V/d

Where;

V is voltage

d is distance

However, the net electric field produced is given by;

E' = V/2d

The tension in the cable can then be expressed as;

F = qE'

Where q is charge

Thus;

F = qV/2d - - - (eq 1)

We also know that;

C = q/V = E_o•A/d

A is area = πr²

Thus;

q/V = E_o•πr²/d

q = VE_o•πr²/d

Let's put VE_o•πr²/d for q in eq 1 to get;

F = V²E_o•πr²/2d²

B) formula for the energy stored in the electric field is;

U = ½CV²

From earlier, we saw that; C = E_o•A/d

Thus;

U = ½E_o•AV²/d

A = πr²

Thus;

U = E_o•Aπr²V²/2d

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nah.   i'm just sleepy.