Two 1.7 cm -diameter disks face each other, 2.5 mm apart. They are charged to ±2

Question

Two 1.7 cm -diameter disks face each other, 2.5 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.

This is the correct answer for A

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.

I've have tried many ways to get the answer to B but every way I try the computer says I'm wrong. Can someone show me step my step how to get the answer to B please.?    Thank you!

Explanation / Answer

The electric field between the plates is

E = /o... where = area charge distribution
= 10X10^-9/(*(0.0085m)^2) = 4.40x10^-5C/m^2
and o = 8.854x10^-12C^2/N-m^2

So E = 4.40x10^-5C/m^2/8.854x10^-12C^2/N-m^2 = 4969505 N/C


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 = 4969505N/C *0.0085m = 42240V

Since Kt = 0 we have Kb = q*E*d = 1.60x10^-19C*42240=6.75x10^-15J

So 1/2*m*v^2 =6.75x10^-15J


Therefore v = sqrt(2*6.75x10^-15J/m) = sqrt(2*6.75x10^-15J/1.67x10^-27kg) = 2843208.106 m/s

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