Question

Question:An electron is accelerated horizontally from rest by a potential difference of 2200 V. It then passes between two horizontal plates 6.5 cm long and 1.3 cm apart that have a potential difference of 250 V (Fig. 17–50). At what angle$\theta$will the electron be traveling after it passes between the plates?

Short answer

The angle at which the electron will be traveling is$$\theta ={15.85}^{\circ }$$.

Step-by-step solution

Understanding of force acting on a point charge in an electric field

In an electric field, the force acting on a charged particle relies on the magnitude of charge and electric field strength.

The force on a charge is given by,

$F=qE$

Here, q is the charge and E is the electric field.

Given Data

The potential difference is,$V=2200\mathrm{V}$.

The length of the horizontal plate is,$L=6.5\mathrm{c}\mathrm{m}$.

The separation distance is,$d=1.3\mathrm{c}\mathrm{m}$.

The potential difference of plates is $V_{\mathrm{p}}=250\mathrm{V}$.

Determination of the electric force and velocity of the electron 

The electric forceon a charge in an electric field is given by,

$\begin{array}{c}F=qE\\ ma=qE\\ m\left(\frac{v_{\mathrm{f}\mathrm{y}}-v_{\mathrm{i}}}{t}\right)=qE\end{array}$

Here, $v_{\mathrm{i}}$ and $v_{\mathrm{f}\mathrm{y}}$ are the initial velocity of election whose value is zero and the vertical component of final velocity of the electron, Eis the electric field, mis the mass, ais the acceleration, tis the time and qis the charge.

Solve the above expression for $v_{\mathrm{f}\mathrm{y}}$.

$\begin{array}{c}m\left(\frac{v_{\mathrm{f}\mathrm{y}}-0}{t}\right)=qE\\ v_{\mathrm{f}\mathrm{y}}=\frac{qEt}{m}\\ v_{\mathrm{f}\mathrm{y}}=\frac{qEL}{m{v}_{\mathrm{f}\mathrm{x}}}\end{array}$

Here, $v_{\mathrm{f}\mathrm{x}}$is the horizontal component of the velocity of the electron.

Determination of the angle made by the electron           

From the conservation of energy,

$\begin{array}{c}PE=KE\\ qV=\frac{1}{2}m{v_{\mathrm{f}\mathrm{x}}}^2\\ 2qV=m{v_{\mathrm{f}\mathrm{x}}}^2\end{array}$

The relation to find angle is given by,

$\begin{array}{c}\tan \theta =\frac{v_{\mathrm{f}\mathrm{y}}}{v_{\mathrm{f}\mathrm{x}}}\\ \tan \theta =\left(\frac{\frac{qEL}{m{v}_{\mathrm{f}\mathrm{x}}}}{v_{\mathrm{f}\mathrm{x}}}\right)\\ \tan \theta =\left(\frac{qEL}{m{v_{\mathrm{f}\mathrm{x}}}^2}\right)\end{array}$

Substitute the values in the above expression.

$$\begin{array}{c}\tan \theta =\left(\frac{q\left(\frac{V_0}{d}\right)L}{2qV}\right)\\ \tan \theta =\left(\frac{\left(\frac{V_0}{d}\right)L}{2V}\right)\end{array}$$

Substitute the values in the above expression.

$$\begin{array}{c}\tan \theta =\frac{\left(\frac{250\mathrm{V}}{1.3\mathrm{c}\mathrm{m}\times \frac{{10}^{-2}\mathrm{m}}{1\mathrm{c}\mathrm{m}}}\right)\left(6.5\mathrm{c}\mathrm{m}\times \frac{{10}^{-2}\mathrm{m}}{1\mathrm{c}\mathrm{m}}\right)}{2\left(2200\mathrm{V}\right)}\\ \theta ={15.85}^{\circ }\end{array}$$

Thus, the angle at which the electron will be traveling is $$\theta ={15.85}^{\circ }$$.