Two 20.1 cm -diameter disks face each other, 1.0 mm apart. They are charged to ±

Question

Two 20.1  cm -diameter disks face each other, 1.0 mm apart. They are charged to ± 1.95  C . Note: In calculating speeds below, please use Newtonian methods, not special relativity. However, you should know that if you do calculate a speed greater than one tenth of the speed of light, in such a case your Newtonian methods are no longer producing accurate results; in those cases, special relativity is required for an reliable prediction of physical reality.

Part A

What is the electric field strength between the disks? Hint: the separation distance is small compared to the radius of the disks, therefore you may use the approximation of infinite sheets of charge.

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?

Part C

An electron is shot from the positive disk toward the negative disk. What launch speed must the electron have to just barely reach the negative disk?

Explanation / Answer

a)

E = Q / X A,

where A = D2/ 4

E= 4 Q / D2

=4 X 1.95 X 10-9 / X 8.854 X 10-12 X 0.22

E = 7.8 X 10-9 / 1.11 X 10-12

E = 7.02 X 103 V/m

b) charge e =1.6 X 10-19 C

proton mass m=1.6726 X 10-27 , d=0.0010 m

Ekinetic energy on negativ disc = Epotential energy on positive disc
1/2 mv2 = eV,

where V= E X d
v2 = e V / 2 m

= (2edE/m)
= (2 X 1.6 X 10-19 X 0.0010 X 7.02 X 103 /1.6 X 10-27)

= (0.022 X 10-16 / 1.6 X 10-27 )

= ( 0.014 X 1011 )

= 34785.05 m/s

= 34.785 km / s

C)

part c answer is also same as above

becuase electron charge and poton charge magnitudes are equal

and the masses of both are equal

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.