C-level 1 A parallel plate capacitor with 2 280cm plates and a 0.76mm thick diel

Question

C-level 1 A parallel plate capacitor with 2 280cm plates and a 0.76mm thick dielectric is found to have a capacitance of 732pF .

a. What is the dielectric constant of the dielectric?

b. Connecting the capacitor to 9 volts DC (our capacitance meter), how much charge is there on the plate connected to the positive side of the power supply?

c. How much charge is there on the plate connected to the negative side of the power supply?

d. From the voltage, calculate the electric field between the plates.

e. What is the charge per unit area on each plate?

f. From the surface charge density, calculate the electric field. Compare to answer in part d.

g. Calculate the energy stored on the capacitor.

h. From the energy stored on the capacitor, calculate the energy density in the dielectric.

i. Calculate the energy density in the dielectric from your value of the electric field.

Explanation / Answer

here,

side of plate , a = 0.0228 m

sepration of plates , d = 0.76 * 10^-3 m

capacitance , C = 0.732 * 10^-9 F

a)

let the dielectric constant be k

k * Area * e0 /d = C

k*0.0228^2 * 8.85 * 10^-12 /( 0.76 * 10^-3) = 0.732 * 10^-9

k = 121

b)

V = 9 V

charge , Q = C * V

Q = 0.732 * 10^-9 * 9

Q = 6.59 * 10^-9 C

the charge on the plate connected to the positive side of the power supply is 6.59 * 10^-9 C

C)

the charge on the plate connected to the negative side of the power supply is - 6.59 * 10^-9 C

D)

electric feild , E = V/d

E = 9 /( 0.76 * 10^-3)

E = 1.18 * 10^4 N/C

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.