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Physics Principles with Applications
Found in: Page 473
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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 \[\theta = 15.85^\circ \].

Most popular questions for Physics Textbooks

(III) A \({\bf{2}}{\bf{.50}}\;{\bf{\mu F}}\) capacitor is charged to 746 V and a \({\bf{6}}{\bf{.80}}\;{\bf{\mu F}}\) capacitor is charged to 562 V. These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? [Hint: Charge is conserved.]

(II) An electron starts from rest 24.5 cm from a fixed point charge with \({\bf{Q = - 6}}{\bf{.50}}\;{\bf{nC}}\). How fast will the electron be moving when it is very far away?

(II) To get an idea how big a farad is, suppose you want to make a 1-F air-filled parallel-plate capacitor for a circuit you are building. To make it a reasonable size, suppose you limit the plate area to . What would the gap have to be between the plates? Is this practically achievable?

(I) The two plates of a capacitor hold \({\bf{ + 2500}}\;{\bf{\mu C}}\) and \( - {\bf{2500}}\;{\bf{\mu C}}\) of charge, respectively, when the potential difference is 960 V. What is the capacitance.

In an older television tube, electrons are accelerated by thousands of volts through a vacuum. If a television set were laid on its back, would electrons be able to move upward against the force of gravity? What potential difference, acting over a distance of 2.4 cm, would be needed to balance the downward force of gravity so that an electron would remain stationary? Assume that the electric field is uniform.
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