Two 2.1cm diameter disks face each other, 2.9mm apart. They are charged to ±10nC. a)...

Ikunupe6v

Ikunupe6v

Answered

2022-01-16

Two 2.1cm diameter disks face each other, 2.9mm apart. They are charged to ±10nC.
a) what is the electric field strength between two disks?
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?

Answer & Explanation

Annie Gonzalez

Annie Gonzalez

Expert

2022-01-17Added 41 answers

Step 1
Given:
Diameter of disk =2.1cm
Radius of disk =1.05cm
=1.05×102m
Distance between two disks, s=2.9mm
=2.9×103m
Charge on disks =±10nC
=±10×109C
Permitivity of free space, ξ0=8.854×1012Fm
Step 2
(a) Electric field strength can be calculated as
E=σξ0
Where, σ is the surface charge density
σ=QA
Where A is the area
A=πr2
Therefore,
E=Qξ0πr2
=10×109C8.854×1012×(1.05×102)2
=3.26×106NC
Step 3
Electric force =qE
Also, Force=ma
Since, both the force experienced by the same charge particle
Therefore, qE=ma
a=qEm
where m is the mass of the proton
m=1.67×1027kg
and e is the charge of the proton
a=1.6×1019×3.26×1061.673×1027
a=3.12×1014ms2
Step 4
Using the equation of motion
s=ut+0.5at2
initial velocity of the particle, u=0
s=0.5at2
t=2sa
=2×2.9×1033.12×1014
=4.3×109s
Step 5
(b) Speed of the proton
Again using the equation of motion
v=u+at
v=0+3.12×1014×4.3×109
=13.42×105ms
Step 6
Answer;
(a) Electric field strength between two disks =3.26×106NC
(b) launch speed of the proton to just barely reach the positive disk =13.42×105ms

Karen Robbins

Karen Robbins

Expert

2022-01-18Added 49 answers

Answer:
E=2.88106NC
v=0.742106ms
Explanation:
The two disks form a parallel-plate capacitor, which will cause an electric field equal to:
E=Qe0A
Where Q is the charge of the disks, A is the area of the disks and e0 is the vacuum permisivity equal to 8.851012C2Nm2:
E=8109C(14π(0.02m)2)8.851012C2Nm2=2.88106NC
Now, for the second part of the problem, we can use conservation of energy. The addition of potential and kinetic energy at launch point should be equal to the addition at the positive disk. Because the proton has positive charge, the potential energy of the proton will increase as its distance to the negative disk increases too. This is because the proton will be attracted towards the negative disk. The potential energy is given by:
Ep=Vq
Where V is the difference in potential (voltage) between the disks. In a parallel-plate capacitor:
V=Ed, where d is the difference in position with the frame of reference. Our frame of reference will be the negative disk.
q is the charge of the proton.
The kinetic energy is given by:
Ek=12mv2
Then:
Ep1+Ek1=Ep2+Ek2
V1q+12mv12=V2q+12mv22v2=0,Ep1=0
12mv12=Edq
v=2Edqm=22.88106Nm0.001m1.61019C1.671027=0.742106ms
alenahelenash

alenahelenash

Expert

2022-01-23Added 366 answers

(A) electric field between to charged disk =E=QAξo Q=charge on each disk=9.0nC r=disk radius=1.5e02m A=surface area of each disk=πr2=0.0007069m2 ξ0=8.854×1012C2/N m2 E=QAξo=1.438e+06NC (B) kinetic energy required by the proton to move from negative disk to positive disk =K K=qV=qEd q=1.602e19C E=1.438e+06NC d=gap between disk=1.0mm K=2.304e016J if the launch speed =v K=mv22; mass of proton =m=1.673e027kg v=2Km=5.249e+05ms

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?