Ikunupe6v

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?

Annie Gonzalez

Expert

Step 1
Given:
Diameter of disk $=2.1cm$
Radius of disk $=1.05cm$
$=1.05×{10}^{-2}m$
Distance between two disks, $s=2.9mm$
$=2.9×{10}^{-3}m$
Charge on disks $=±10nC$
$=±10×{10}^{-9}C$
Permitivity of free space, ${\xi }_{0}=8.854×{10}^{-12}\frac{F}{m}$
Step 2
(a) Electric field strength can be calculated as
$E=\frac{\sigma }{{\xi }_{0}}$
Where, $\sigma$ is the surface charge density
$\sigma =\frac{Q}{A}$
Where A is the area
$A=\pi {r}^{2}$
Therefore,
$E=\frac{Q}{{\xi }_{0}\pi {r}^{2}}$
$=\frac{10×{10}^{-9}C}{8.854×{10}^{-12}×{\left(1.05×{10}^{-2}\right)}^{-2}}$
$=3.26×{10}^{6}\frac{N}{C}$
Step 3
Electric force $=qE$
Also, $\text{Force}=ma$
Since, both the force experienced by the same charge particle
Therefore, $qE=ma$
$a=\frac{qE}{m}$
where m is the mass of the proton
$m=1.67×{10}^{-27}kg$
and e is the charge of the proton
$a=\frac{1.6×{10}^{-19}×3.26×{10}^{6}}{1.673×{10}^{-27}}$
$a=3.12×{10}^{14}\frac{m}{{s}^{2}}$
Step 4
Using the equation of motion
$s=ut+0.5a{t}^{2}$
initial velocity of the particle, $u=0$
$s=0.5a{t}^{2}$
$t=\sqrt{\frac{2s}{a}}$
$=\sqrt{\frac{2×2.9×{10}^{-3}}{3.12×{10}^{14}}}$
$=4.3×{10}^{-9}s$
Step 5
(b) Speed of the proton
Again using the equation of motion
$v=u+at$
$v=0+3.12×{10}^{14}×4.3×{10}^{-9}$
$=13.42×{10}^{5}\frac{m}{s}$
Step 6
(a) Electric field strength between two disks $=3.26×{10}^{6}\frac{N}{C}$
(b) launch speed of the proton to just barely reach the positive disk $=13.42×{10}^{5}\frac{m}{s}$

Karen Robbins

Expert

$E=2.88\cdot {10}^{6}\frac{N}{C}$
$v=0.742\cdot {10}^{6}\frac{m}{s}$
Explanation:
The two disks form a parallel-plate capacitor, which will cause an electric field equal to:
$E=\frac{Q}{{e}_{0}A}$
Where Q is the charge of the disks, A is the area of the disks and ${e}_{0}$ is the vacuum permisivity equal to $8.85\cdot {10}^{-12}\frac{{C}^{2}}{N}{m}^{2}$:
$E=\frac{8\cdot {10}^{-9}C}{\left(\frac{1}{4}\pi {\left(0.02m\right)}^{2}\right)\cdot 8.85\cdot {10}^{-12}\frac{{C}^{2}}{N}{m}^{2}}=2.88\cdot {10}^{6}\frac{N}{C}$
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:
${E}_{p}=V\cdot q$
Where V is the difference in potential (voltage) between the disks. In a parallel-plate capacitor:
$V=E\cdot d$, 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:
${E}_{k}=\frac{1}{2}m{v}^{2}$
Then:
${E}_{p1}+{E}_{k1}={E}_{p2}+{E}_{k2}$
${V}_{1}\cdot q+\frac{1}{2}m{v}_{1}^{2}={V}_{2}\cdot q+\frac{1}{2}m{v}_{2}^{2}\mid {v}_{2}=0,{E}_{p1}=0$
$\frac{1}{2}m{v}_{1}^{2}=E\cdot d\cdot q$
$v=\sqrt{\frac{2Edq}{m}}=\sqrt{\frac{2\cdot 2.88\cdot {10}^{6}\frac{N}{m}\cdot 0.001m\cdot 1.6\cdot {10}^{-19}C}{1.67\cdot {10}^{-27}}}=0.742\cdot {10}^{6}\frac{m}{s}$

alenahelenash

Expert

(A) electric field between to charged disk $=\stackrel{\to }{E}=\frac{Q}{A{\xi }_{o}}$ $Q=\text{charge on each disk}=9.0nC$ $r=\text{disk radius}=1.5e-02m$ $A=\text{surface area of each disk}=\pi {r}^{2}=0.0007069{m}^{2}$ $⇒\stackrel{\to }{E}=\frac{Q}{A{\xi }_{o}}=1.438e+06\frac{N}{C}$ (B) kinetic energy required by the proton to move from negative disk to positive disk $=K$ $K=q\mathrm{△}V=qEd$ $q=1.602e-19C$ $E=1.438e+06\frac{N}{C}$ $d=\text{gap between disk}=1.0mm$ $⇒K=2.304e-016J$ if the launch speed $=v$ $K=\frac{m{v}^{2}}{2}$; mass of proton $=m=1.673e-027kg$ $v=\sqrt{\frac{2K}{m}}=5.249e+05\frac{m}{s}$