Question

Two 2.2-cm-diameter-disks spaced 1.8mm apart form a parallel-plate capacitor. The electric field between the disks is 4.1

Answer 1

Answer 2

A) V = E*d

= 4.1*10^5*1.8*10^-3

= 738 volts <<<<<<<-----------Answer

B) C = A*epsilon/d

= pi*r^2*epsilon/d

= pi*0.011^2*8.854*10^-12/1.8*10^-3

= 1.87*10^-12 F

charge on positive plate, q1= C*V

= 1.87*10^-12*738

= 1.38*10^-9 C <<<<<<<-----------Answer

charge on neagative plate, q2 = -1.38*10^-9 C <<<<<<<-----------Answer

C) Apply, F = m*a

q*E = m*a

a = q*E/m

= 1.6*10^-19*4.1*10^5/9.1*10^-31

= 7.2*10^16 m/s^2

let u is the initail speed of electron at negtaive plate.

now, Apply v^2 - u^2 = 2*a*s

u^2 = v^2 - 2*a*s

u = sqrt(v^2 - 2*a*s)

= sqrt((2.1*10^7)^2 - 2*7.2*10^16*1.8*10^-3)

= 1.348*10^7 m/s <<<<<<<-----------Answer

Answer 3

A. Since the electric eld is constant inside the capacitor, the potential is just V = Ed, where d is the separation distance.

V = 4.1*10^5*1.8*10^-3 = 738 V

B.

The electric field between the capacitor is E = η/60 where η = Q/A the surface charge density. So, Q-: πR2E0E. Thus, n/Eg, where η Q/rR9s

Q = 3.14*(1.1*10^-2)^2*8.85*10^-12*4.1*10^5 = 1.38*10^-9 C

C.

,2-2oV

Vi = [(2.1*10^7)^2 - [2*1.6*10^-19*738/(9.11*10^-31)]]^0.5 = 1.348*10^7 m/s