A parallel-plate capacitor is made from two plates 11.0 cm on each side and 4.65

Question

A parallel-plate capacitor is made from two plates 11.0 cm on each side and 4.65 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40. (See the figure below(Figure 1).) An 20.0 V battery is connected across the plates.

a;What is the capacitance of this combination?

b; How much energy is stored in the capacitor?

c; If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor?

Explanation / Answer

a)

Work this as two capacitors in parallel. each one has half the area, or an area of

(11x11)/2 = 60.5 cm² or 60.5e-4 m²

Parallel plate cap

C = (A/d) in Farads

is 8.854e-12 F/m

is dielectric constant (vacuum = 1)

A and d are area of plate in m² and separation in m

The air one, C = (8.854e-12)(60.5e-4) /(0.00465) = 1.152 x 10^-11 F

The other, C = (8.854e-12)(3.4)(60.5e-4) /(0.00465) = 3.917 x 10^-11 F

add the two for the total C

Total C = (1.152 x 10^-11) + (3.917 x 10^-11)

= 5.069 x 10^-11 F

b)

E = ½ * C* V^2

= ½ * 5.069 x 10^-11 *20^2

= 1.014 x 10^-8 J

c)

Now,

A = (11x11) = 121 cm² or 121e-4 m²

C = (8.854e-12)(121e-4) /(0.00465) = 2.304 x 10^-11 F

Thus,

E = ½ * 2.304 x 10^-11 * 20^2

= 4.608 x 10^-9 J

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