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An electron is placed at the centre of an ideal parallel plate capacitor where the plates are separated by $d$ metres, and the potential difference between the plates is $V$ Volts. The orientation and arrangement of the component parts including the electron is indicated in the diagram above.  The upper electrode is at the more positive electrostatic potential.

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In your solution you should replace

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$\\displaystyle{1\\over{4\\pi\\varepsilon}}$

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with $k$, and denote the charge on an electron as $-q$.  When providing numerical answers you may express them using scientific notation.  Express values to four significant figures and use the values of physical constants as provided in the course notes.

From Gauss' Law for an infinite parallel plate capacitor we know that the electric field, $E=V/d$, is uniform in both magnitude and direction between the plates, and points from the positive electrode to the negative electrode.  The electron generates a spherically symmetric field (Coulomb's Law), which can be viewed as

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$\\displaystyle{-e\\over 4 \\pi \\varepsilon_0 r^2}$

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$\\displaystyle{e\\over 4 \\pi \\varepsilon_0 r^2}$

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The field from the capacitor and the field from the electron add vectorially everywhere, so that midway between the electron and the positive electrode, where both fields are 'down', they combine to increase the field, giving

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$\\displaystyle{{V\\over d}+{e\\over 4 \\pi \\varepsilon_0 (d/4)^2}}$ downwards.

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The $d/4$ arises since this point is 1/4 of the way between the electrodes, the electron being located half way between them.  For halfway between the electron and the negative electrode the fields point in opposite directions and tend to cancel, yielding

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$\\displaystyle{{V\\over d}-{e\\over 4 \\pi \\varepsilon_0 (d/4)^2}}$ downwards.

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In practice the field from the capacitor will be generally be vastly bigger than the contribution from the electron.

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Since the electron is negatively charged it will experience a force $\\vec{F}=q\\vec{E}=-e\\vec{E}$ which is upwards.  You were asked to calculate the magnitude in femto-Newtons, so you need to remember that 'femto' means '$10^{-15}$'.  Note that the area is given in the question but is not needed to perform any calculation - in real life you often have lots of information that does not help you solve the problem, and often you don't have some of the information that you really want.

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Energy stored in J

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Force on the electron in N

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Charge denisity on each plate in C/m^2

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PPC area in mm^2.

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Charge on each plate

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Potential difference across plates in Volts.

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Force on the electron in femto Newtons

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Plate separation in micrometres.

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Separation in metres.

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Capacitance of device in F

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Permittivity of free space in F/m

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E field (uniform) in V/m

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Magnitude of the charge on a electron in C.

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When combining two or more sources of electric field we should use...

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Write down the formula for the field midway between the positive plate and the electron.  Express your answer in terms of $V$, $d$, $k$ and $q$.  Do not use 'e' to represent the charge on a proton, as this will be interpretted as exp(1).

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[[0]] in the downward direction on the diagram

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Write down the formula for the field midway between the negative plate and the electron.  Express your answer in terms of $V$, $d$, $k$ and $q$.   Again, do not use 'e' for the charge on a proton as this will be interpretted as exp(1).

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[[0]] in the downward direction on the diagram

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The capacitor is biased so that the potential difference across the plates is {pd} Volts. The distance between the plates is {wmicron} μm and the area of each plate is {areamm} mm$^2$. What force does the electron experience?

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$|\\vec{F}|=$ [[0]] femto-Newtons in the [[1]] direction.

This means that the electron is moving in the direction of [[0]] $E$-field.
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