Two 3.0-cm-diameter disks face each other, 1.0 mm apart. They are charged to ± 1

Question

Two 3.0-cm-diameter disks face each other, 1.0 mm apart. They are charged to ± 10 nC .

Part A

What is the electric field strength between the disks?

Express your answer using two significant figures.

3.6•106

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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 using two significant figures.

Two 3.0-cm-diameter disks face each other, 1.0 mm apart. They are charged to ± 10 nC .

Part A

What is the electric field strength between the disks?

Express your answer using two significant figures.

E =

3.6•106

  N/C  

SubmitMy AnswersGive Up

Incorrect; Try Again; 9 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 using two significant figures.

v =   m/s  

Explanation / Answer

The electric field between the plates is

E = /o   where = area charge distribution

= 10X10-9/(*(0.010)2) = 3.183x10-5 C/m2
and o = 8.854x10-12 C2/N-m^2

So E = 3.183x10-5 /8.854x10-12 = 3.60x106N/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 = 3.60x10^6V/m*1.0x10^-3m = 3595V

Since Kt = 0 we have Kb = q*E*d = 1.60x10-19 *3595 = 5.75x10-16J

So 1/2*m*v2 = 5.75x10-16J


Therefore v = sqrt(2*5.75x10-16J/m) = sqrt(2*5.75x10^-16 /1.67x10^-27kg) = 8.30x105m/s

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