A proton is fired from the low potential disk toward the high potential disk. At what speed the proton will barely reach the high potential disk?
This question aims to explain electric field strength, electric charge, surface charge density, and equation of motion. The electric charge is the characteristic of subatomic particles that compels them to encounter a force when held in an electric and magnetic field whereas an electric field is defined as the electric force per unit charge. The formula of the electric field is:
E = FQ
Surface charge density $(\sigma)$ is the amount of charge per unit area, and equations of motion of kinematics define the basic idea of the motion of a thing such as the position, velocity, or acceleration of a thing at different times.
Expert Answer
Part A:
Data given in the question is:
- Diameter of the disk $d = 2.1cm$
- Radius of the disk $r=\dfrac{2.1}{2} = 1.05cm$ = $1.05 \times 10^{-2} m$
- Distance between the disks, $s = 2.9mm$ = $2.9 \times 10^{-3}$
- Charge on the disks $Q= \pm 10nC$ = $ \pm 10 \times 10^{-9} C$
- Permittivity of the free space $\xi_o = 8.854 \times 10^{-12} \space F/m$
We are asked to find the Electric Field strength. The formula for Electric Field strength is given as:
\[E = \dfrac{\sigma}{\xi}\]
Where the $\sigma$ is surface charge density and is given as:
\[\sigma=\dfrac{Q}{A}\]
$A$ is the area given by $\pi r^2$.
Electric Field strength $E$ can be written as:
\[E = \dfrac{Q}{\xi \pi r^2}\]
Plugging the values:
\[E = \dfrac{10 \times 10^{-9} C}{(8.854 \times 10^{-12}) \pi (1.05 \times 10^{-2})^2 }\]
\[ 3.26 \times 10^{6} N/C \]
Part B:
Since the Electrical force $F=qE$ and the force $F=ma$ experience the same charge particle, therefore:
\[qE=ma\]
\[a=\dfrac{qE}{m}\]
- $m$ is mass of proton that is $1.67 \times 10^{-27} kg$
- $q$ is the charge of proton that is $1.6 \times 10^{-19}$
Inserting values into the formula:
\[a= \dfrac{(1.6 \times 10^{-19})(3.26 \times 10^{6})}{1.67 \times 10^{-27}}\]
\[a= 3.12 \times 10^{14} m/s\]
Using the equation of motion to calculate the time:
\[s = ut+0.5at^2\]
Where the initial velocity $u$ is $0$.
\[s = 0.5at^2\]
\[t= \ \sqrt{\dfrac{2s}{a}}\]
Inserting the values:
\[t= \ \sqrt{\dfrac{(2.9 \times 10^{-3})}{ 3.12 \times 10^{14}}} \]
\[ t = 4.3 \times 10^{-9}s \]
For calculating the speed of the proton, equation of motion is used as:
\[v = u + at\]
Inserting the values to calculate the $v$.
\[ v = 0 + (3.12 \times 10^{14}) (4.3 \times 10^{-9}) \]
\[ v = 13.42 \times 10^5 m/s \]
Numerical Answer
Part a: $E$ between two disks is $3.26\times 10^{6} N/C$.
Part b: The launch speed is $13.42 \times 10^5 m/s$.
Example
Specify the magnitude of the electric field $E$ at a point $2cm$ left of a point charge of $−2.4 nC$.
\[E= k\dfrac{q}{r^2} \]
\[E = k\dfrac{(9\times 10^9)(2.4\times 10^{-9})}{0.02^2} \]
\[E = 54\times 10^3 N/C \]
In this problem, the charge is negative $−2.4 nC$, so the direction of the electric field will be towards that charge.