**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 w**hereas 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.**