Question

The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. The potential difference between the plates is 25.0 V. Typical dimensions are about 3.3 cm on a side, with a separation of about 5.2 mm. The plates are close enough that we can ignore fringing at the ends. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

Answer 1

(a) $4.68\cdot 10^{-11}C$

The capacitance of a parallel plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0 = 8.85 \cdot 10^{-12} F/m$ is the vacuum permittivity

A is the area of each plate

d is their separation

For the capacitor in the problem:

$A=(0.033m)^2=0.0011 m^2$

d = 5.2 mm = 0.0052 m

Substituting,

$C=\frac{(8.85\cdot 10^{-12})(0.0011)}{0.0052}=1.87\cdot 10^{-12} F$

The capacity is related to the charge on the plate by:

$Q=CV$

where

V = 25.0 V is the potential difference between the plates

Substituting,

$Q=(1.87\cdot 10^{-12})(25.0)=4.68\cdot 10^{-11}C$

(b) 4808 V/m

The magnitude of the electric field between the plates is given by:

$E=\frac{V}{d}$

where

V is the potential difference

d is the separation between the plates

Substituting:

V = 25.0 V

d = 5.2 mm = 0.0052 m

We find:

$E=\frac{25.0}{0.0052}=4808 V/m$

(c) $2.96\cdot 10^6 m/s$

When the electron moves from the negative plate to the positive plate, its electric potential energy is converted into kinetic energy, according to the equation:

$qV=\frac{1}{2}mv^2$

where

$q=1.6\cdot 10^{-19} C$ is the magnitude of the electron's charge

V is the potential difference

$m=9.1\cdot 10^{-31} kg$ is the mass of the electron

v is the final speed of the electron

Solving for v, we find:

$v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19})(25.0}{9.1\cdot 10^{-31})}}=2.96\cdot 10^6 m/s$