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A parallel-plate capacitor has square plates that are 9.00 cmcm on each side and 4.10 mmmm apart. The space between the plates is completely filled with two square slabs of dielectric, each 9.00 cmcm on a side and 2.05 mmmm thick. One slab is pyrex glass and the other is polystyrene.

If the potential difference between the plates is 86.0 V, how much electrical energy is stored in the capacitor?

User TrapII
by
5.4k points

2 Answers

2 votes

Answer:

2.29 x 10^-7 J.

Step-by-step explanation:

Hiven:

L = 9 cm

= 0.09 m

d = 2.05 mm

= 2.05 x 10^-3 m

ε° * kpo = 2.56 * 8.854 x 10^-12

= 2.27 x 10-11

ε° * kpo = 5.6 * 8.854 x 10^-12

= 5 x 10-11

V = 86 V

The two plates of polystyrene and pyrex glass as a two separate capacitors connected on series so the total capacitance of series capacitor is given by:

1/CT = 1/Cpy + 1/Cpo

Capacitance of polystyrene, Cpo = ε° * kpo * A/d

Area = L^2

= 0.09^2

= 0.0081 m^2

d = 0.00205 m

Cpo = 2.27 x 10^-11 * 0.0081/0.00205

= 9 x 10^-11 F

Capacitance of pyrex, Cpy = ε° * kpy * A/d

= 5 x 10^-11 * 0.0081/0.00205

= 2 x 10^-10 F

1/CT = 1/9 x 10^-11 + 1/2 x 10^-10

CT = 6.19 x 10^-11 F

Energy stored in the capacitor

U = 1/2 * C * V^2

= 1/2 * 6.19 x 10^-11 * 86^2

= 2.29 x 10^-7 J.

User Adrien Delessert
by
5.4k points
6 votes

Answer:

U= 1.9*10^-7

Step-by-step explanation:

given:

L=8 cm

d_1=d_2=2.05_10^-3

K_py=4.3

K_po=3

V_ab=86 V

required

U=??

solution:

the energy stored in the capacitor

U=1/2C_t*V^2_ab (1)

voltage is known but capacitance is not

we can consider the two plates of polystyrene and pyrex glass as a two separate capacitors connected on series so the total capacitance of series capacitor is given by:

1/C_t= 1/C_py+1/C_po (2)

the capacitance of polystyrene

C_po=K_po*A/d

=8.9*10^-11

= 89 pF

the capacitance of pyrex

C_py=K_py*A/d

=128 pF

substitution in 2 yields

1/C_t= 1/128+1/89

C_t= 52.6 pF

substitution in 1 yields

U=1/2C_t*V^2_ab

= 1.9*10^-7

User Jlecour
by
5.4k points