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93GP

Expert-verifiedFound in: Page 473

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

**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.

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 \({V_{\rm{p}}} = 250\;{\rm{V}}\).

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 *q*is 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.

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

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