24.41 A capacitor has parallel plates of area 12 cm2 separated by 4.0 mm . The s

Question

24.41 A capacitor has parallel plates of area 12 cm2 separated by 4.0 mm . The space between the plates is filled with polystyrene (K=2.6, Em=2.0×107V/m).

1. Find the permittivity of polystyrene.

2. Find the maximum permissible voltage across the capacitor to avoid dielectric breakdown.

3. When the voltage equals the value found in part B, find the surface charge density on each plate.

4. Find the induced surface-charge density on the surface of the dielectric.

Do show your steps please. I would like to understand. Thank you! :)

Explanation / Answer

A capacitor has parallel plates of area 12 cm2 separated by 4.0 mm .

The space between the plates is filled with polystyrene (K=2.6, Em=2.0×107V/m).

1)

the permittivity of polystyrene.

K of polysterene = 2.6

--------------------------------------------

2)

apply k = V/Vo

also Electric field E = V/d

so

Vo = Ed

Vo = 2*10^7 * 4*10^-3

Vo = 80 KVolts

so

the maximum permissible voltage is

Vmax = 80*2.6   = 208 KVolts

--------------------------------------------

3)

surface charge density sigma = Q/A = E eo

sigma = 2e7 * 8.85*10^-12

the voltage equals the value found in part B, find the surface charge density on each plate is

sigma = 1.77 *10^-4 C/m^2

--------------------------------------------

4)

E = Eo/k    = 2e7/2.6 = 7.69 e 6 V/m

so

sigma = Eeo = 7.69*10^6 * 8.85*10^-12

the induced surface-charge density on the surface of the dielectric is

sigma = 6.733 *10^-5 C/m^2

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