An air-gap parallel-plate capacitor is constructed with plates that have an area

Question

An air-gap parallel-plate capacitor is constructed with plates that have an area of 0.25 m2 and plate-separation of 0.15 mm. The capacitor is charged so that the two plates each have a surface charge density of magnitude 7.5 µC/m2 .

(a) Find the electric field between the two plates and use your value to determine the voltage on the capacitor.

(b) Find the capacitance of the capacitor and the charge on each of its plates.

(c) If a proton is released from rest at the positive plate, what will be its velocity when it hits the negative plate?

(d) If a dielectric with a dielectric constant = 4.9 is inserted in between the plates, what is the new voltage on the capacitor?

(e) If the dielectric strength of the material is 28 × 106 V/m, find the maximum charge that can be placed on the capacitor

3. An air-gap parallel-plate capacitor is constructed with plates that have an area of 0.25 m2 and plate-separation of 0.15 mm. The capacitor is charged so that the two plates each have a surface charge density of magnitude 7.5 uC/m2 (a) Find the electric field between the two plates and use your value to determine the voltage on the capacitor. (b) Find the capacitance of the capacitor and the charge on each of its plates. (c) If a proton is released from rest at the positive plate, what will be its velocity when it hits the negative plate? (d) If a dielectric with a dielectric constant K = 4.9 is inserted in between the plates, what is the new voltage on the capacitor? (e) If the dielectric strength of the material is 28 x 106 V/m, find the maximum charge that can be placed on the capacitor.

Explanation / Answer

Given  

air gap parallel plate capacitor with Area of the plates is  

A = 0.25 m^2

separation of the plates is d = 0.15 mm = 0.15*10^-3 m

surface charge density sigma = 7.5*10^-6 C/m^2

we know that the electric field between the two plates of a parallel place capacitor is  

a)

E = sigma/epsilon not

E = 7.5*10^-6 / (8.854*10^-12) V/m

E = 847074.77 V/m

voltage on each plate is V = E*d = 847074.77*0.15*10^-3 V = 127.06 V

b) capacitance is C = epsilon0*A/d = 8.854*10^-12*0.25/(0.15*10^-3) C

C = 14.75*10^-9 C = 14.75 nC

c) We know that V = W/q = 0.5*m(v^2) /q

v^2 = 2*q*V/m

v = sqrt(2*q*V/m)

v = sqrt(2*1.6*10^-19*127.06/1.6*10^-27) m/s

v = 159411.42 m/s

d) k = 4.9  

if we insert dielectric between plates then C' = k*C0

Q = C*V ==> V = Q/C ==> V' = V/k = 127.06/4.9 v = 25.93 V

e) Q = C*V = C*E*d ==> 14.75*10^-9*28*10^6*0.15*10^-3 C = 6.195*10^-5 c = 61.95*10^-6 C

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