The plates of a parallel plate capacitor are 4 cm wide and 9 cm long. The plates

Question

The plates of a parallel plate capacitor are 4 cm wide and 9 cm long. The plates are separated by a 1.2-mm-thick layer of paper.

(a)

Calculate the capacitance of the device (in pF) using the dielectric constant of paper from this table.

pF

(b)

Any dielectric material other than a vacuum has a maximum electric field that can be generated in the dielectric material before it physically or chemically breaks down and begins to conduct. This maximum electric field is called dielectric strength. The dielectric strength for paper is reached at a value of 15 106 V/m. Calculate the maximum charge (in µC) that can be placed on the capacitor at this dielectric strength.

µC

Explanation / Answer

a) Use equation,

w=0.04m, l=0.09m,d=0.0012m

C0 = 0A/d = 0wl/d= (8.85*10^-12*0.04*0.09)/0.0012 = 2.66*10^-11 F = 26.6*10^-12 F = 26.6pF

When dielectric constant(k) is introduced capacitance increases by factor k

Cd = kdC0 = kd*C0 = kd0A/d

b)

Vd = Ed*d ------------(1)

Qd = CdVd = (kd0A/d)* Ed*d = kd0A Ed

Where Ed = Electric field after dielectric is introduced = 15*10^6 V/m , Cd = capacitance after dielectric is introduced , Qd = charge after dielectric is introduced , kd= dielectric constant of dielectric material here paper

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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