Two circular metal discs each of 3.00 cm. diameter are spaced 1.50 mm apart to c

Question

Two circular metal discs each of 3.00 cm. diameter are spaced 1.50 mm apart to create a capacitor.

(a)What is the capacitance of this capacitor? If this capacitor is connected in series to a 25V battery, (b)What magnitude charge develops on each plate of the capacitor? While the battery remains connected, the plates are pulled apart until the separation between the plates in 2.5mm. (HINT: Think about whether it will remain constant in this scenario) (c)What is the capacitance now? (d) What potential difference is between the plates? (e)What charge develops on each plate of the capacitor?

Now the battery is disconnected and the plates are pushed back to their original 1.50 mm spacing (HINT: Again, think about whether it will be the charge on the plates or the potential across the plates that will remain constant in this scenario as opposed to the above scenario.) (f) What is the capacitance now? (g)What potential difference develops between the plates (h)What charge develops on each plate? (i)What is the magnitude of the electric field between the plates? (j)How much energy is stored in the capacitor? (k)If at this time the capacitor is allowed to discharge through a 15 ohm resistor, what will be the time constant of this RC circuit? And (l) how long will it be before it is half discharged?

Explanation / Answer

a Now that we know Q, we can solve for C. C = Q/V b This question can be answered by (a) assuming that the total stored charge does not change during the experiment, (b) making use of the relationship Q = V*C where Q is charge, V is voltage, C = capacitance (c) when capacitors are connected in parallel, the total capacitance is the sum of their individual capacitances, (d) when components are connected in parallel, the applied voltage is the same for all components. pot. diff (V) = charge (Q) / capacitance (C) = Q * plate separation (d) /(permittivity (e) * Plate Area(A)) E = V/d = Q / e * A C = e0*area/separation V = Q/C E = V/separation = Q/(e0*area) Energy stored in capacitor after dielectric is placed =Uf =(diectric constant)CV^2 = plug and get

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