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Expert-verified Found in: Page 473 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # 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? The angle at which the electron will be traveling is$\theta = 15.85^\circ$.

See the step by step solution

## Step 1: 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.

## Step 2: Given Data

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

The length of the horizontal plate is,$$L = 6.5\;{\rm{cm}}$$.

The separation distance is,$$d = 1.3\;{\rm{cm}}$$.

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

## Step 3: Determination of the electric force and velocity of the electron

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

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

Here, $${v_{\rm{i}}}$$ and $${v_{{\rm{fy}}}}$$ are the initial velocity of election whose value is zero and the vertical component of final velocity of the electron, E is the electric field, m is the mass, a is the acceleration, t is the time and qis the charge.

Solve the above expression for $${v_{{\rm{fy}}}}$$.

$$\begin{array}{c}m\left( {\frac{{{v_{{\rm{fy}}}} - 0}}{t}} \right) = qE\\{v_{{\rm{fy}}}} = \frac{{qEt}}{m}\\{v_{{\rm{fy}}}} = \frac{{qEL}}{{m{v_{{\rm{fx}}}}}}\end{array}$$

Here, $${v_{{\rm{fx}}}}$$is the horizontal component of the velocity of the electron.

## Step 4: Determination of the angle made by the electron

From the conservation of energy,

$$\begin{array}{c}PE = KE\\qV = \frac{1}{2}m{v_{{\rm{fx}}}}^2\\2qV = m{v_{{\rm{fx}}}}^2\end{array}$$

The relation to find angle is given by,

$$\begin{array}{c}\tan \theta = \frac{{{v_{{\rm{fy}}}}}}{{{v_{{\rm{fx}}}}}}\\\tan \theta = \left( {\frac{{\frac{{qEL}}{{m{v_{{\rm{fx}}}}}}}}{{{v_{{\rm{fx}}}}}}} \right)\\\tan \theta = \left( {\frac{{qEL}}{{m{v_{{\rm{fx}}}}^2}}} \right)\end{array}$$

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.

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

Thus, the angle at which the electron will be traveling is $\theta = 15.85^\circ$. ### Want to see more solutions like these? 