Problem 23.26 Two 1.7 cm-diameter disks face each other, 2.2 mm apart. They are

Question

Problem 23.26 Two 1.7 cm-diameter disks face each other, 2.2 mm apart. They are charged to ±19 C. Part A What is the electric field strength between the disks? Express your answer to two significant figures and include the appropriate units. . EValue Units Submit My Answers Give Up 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 My Answers Give Up Provide Feedback Continue

Explanation / Answer

Part A)

electric field between the disks is twice the electric field due to each disk

E = 2*(sigma)/(2*eo) = sigma/(eo)

sigma is the surface charge density = Q/A = (19*10^-9)/(3.142*(0.017/2)^2)) = 8.37*10^-5 C/m^2

eo is the permittivity of free space = 8.85*10^-12 C^2/(N-m^2)

then
E = sigma /eo = (8.37*10^-5)/(8.85*10^-12) = 9.5*10^6 N/C or V/m

B)

Using work energy theorem

Work done on proton = change in kinetic energy

q*E*S = 0.5*m*v^2

q = charge on proton = 1.6*10^-19 C

m is the mass of the proton = 1.67*10^-27 Kg

(1.6*10^-19*9.5*10^6*2.2*10^-3) = 0.5*1.67*10^-27*v^2

v = 2*10^6 m/sec

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