CH 23 HW roblem 23.26 Two 2.4 cm-diameter disks face each other, 1.6 mm apart. T

Question

CH 23 HW roblem 23.26 Two 2.4 cm-diameter disks face each other, 1.6 mm apart. They are charged to -20nC Part A What is the electric field strength between the disks? Express your answer to two significant figures and include the appropriate units 49954.5 Submit X Incorrect; Try Again; 2 attempts remaining Part B A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk? Express your answer to two significant figures and include the appropriate units Value Units Submit Racuost Answor Provide Feedback

Explanation / Answer

A)

The electric field between the plates is

E = /o...

where = area charge distribution

= Q/pi*r^2;   r = 2.4/2 cm = 1.2 cm = 0.012m

Thus

= 20X10^-9/(*(0.012m)^2) = 4.42x10^-5C/m^2

and o = 8.854x10^-12C^2/N-m^2

So E = 4.42x10^-5C/m^2/8.854x10^-12C^2/N-m^2

= 5.0 x 10^6 N/C

B)

Using conservation of energy we solve the second part

(K + U) b = (K + U)t...

Here Ut - Ub = q*(Vt - Vb)...

and Vt - Vb = E*d = 4.99x10^6V/m*1.6x10^-3m =7989V

Since Kt = 0 we have Kb = q*E*d = 1.60x10^-19C*7989 = 1.278 x 10^-15 J

So 1/2*m*v^2 = 1.278 x 10^-15 J

Therefore

v = sqrt(2*1.278x10^-15J/m)

= sqrt(2*1.278x10^-15J/1.67x10^-27kg)

= 1.24 x 10^6 m/s

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