# 12.7.2 Charged Metal Parallel Plates

Consider two large parallel (metal) plates connected across a battery. A current cannot flow across the plates because the air in between is an insulator. Nevertheless, the voltage held between the plates cause excess charges to reside on the surfaces of the plates facing each other. The charged plates produce an electric field in the region between them.

If the plates are large compared to the plate separation d, the field produced is a uniform one. If the plates are not large enough, then the field is very nearly uniform near the centre and less so as we near the edges of the plates (called edge effects). The H2 syllabus assumes a uniform field between the plates and zero elsewhere.

For example, the electric field between two plates held at +6.0 V and −6.0 V is assumed to be uniform. See how the field lines are drawn to be straight and equally spaced out? See how the equipotential lines are drawn perpendicular to the field lines and equally spaced out too? It is all very simple because the field is uniform. Since $\displaystyle E=-\frac{{dV}}{{dx}}$, the field strength between the parallel plates is simply the potential difference between the plates divided by the separation between them.

For this particular one, \displaystyle \begin{aligned}E&=\frac{{\Delta V}}{{\Delta x}}\\&=\frac{{6.0-(-6.0)}}{{4.0\times {{{10}}^{{-3}}}}}\\&=3000\text{ V }{{\text{m}}^{{-1}}}\end{aligned}

The direction of the field is from the positive plate towards the negative plate.

Parallel plates have wide applications where a uniform electric field is needed. For example, in a cathode ray tube (CRT), the electron beam is passed through a pair of parallel plates called the deflection plates. By applying the correct voltage to the plates, the beam can be deflected by the correct amount and made to land at the intended spot on the fluorescent screen.

Example

A beam of electrons, travelling at $\displaystyle 1.35\times {{10}^{7}}\text{ m }{{\text{s}}^{{-1}}}$, passes through a uniform electric field between two plates of length 0.060 m and separation of 0.020 m, before striking the fluorescent screen placed 0.230 m from the centre of the plates. The top plate is at +50 V while the lower plate is at −50 V. Calculate the position where the electron beam strikes the fluorescent screen.

Solution

Electric field strength $\displaystyle E=\frac{{100}}{{0.020}}=5000\text{ V }{{\text{m}}^{{-1}}}$

Acceleration of electrons, $\displaystyle \displaystyle a=\frac{{{{F}_{e}}}}{{{{m}_{e}}}}=\frac{{eE}}{{{{m}_{e}}}}=\frac{{(1.60\times {{{10}}^{{-19}}})(5000)}}{{9.11\times {{{10}}^{{-31}}}}}=8.78\times {{10}^{{14}}}\text{ m }{{\text{s}}^{{-2}}}$

Time taken to pass through the plates, $\displaystyle \Delta t=\frac{{0.060}}{{1.35\times {{{10}}^{7}}}}=4.44\text{ ns}$

Vertical velocity upon exiting the plates, $\displaystyle {{v}_{y}}=0+(8.78\times {{10}^{{14}}})(4.44\times {{10}^{{-9}}})=3.898\times {{10}^{6}}\text{ m }{{\text{s}}^{{-1}}}$

Deflection angle, $\displaystyle \theta ={{\tan }^{{-1}}}(\frac{{3.898\times {{{10}}^{6}}}}{{1.35\times {{{10}}^{7}}}})=16.1{}^\circ$

The apparent origin is at the centre of the deflection plates. $\displaystyle \frac{{\Delta y}}{{0.23}}=\tan 16.1{}^\circ \text{ }\Rightarrow \text{ }\Delta y=6.64\text{ cm}$

Concept Test

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