Name: Forces Between Multiple Charges
Uploaded: 2019-07-19
Duration: 6 min 57 s
Description: To find the force on a charge due to multiple charges by applying the superposition principle, we need to know the basic concepts first. Watch the video to l...

Question

An electron is projected between the plates with an electric field applied with a velocity $v_{x} = 2.0 \times 10^{6} m / s$ as shown in the figure. If the uniform electric field between the plates is $9.1 \times 10^{2} N / C$ and the separation between the plates is $0.5 cm$ , where will the electron strike the upper plate? $(|e| = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg)$

Accepted Answer

The direction of electric field between the plates is as shown in the figure below. Electron is a negatively charged particle it will experience the force due to electric field in a direction opposite to it. The direction of force on it will be towards the upper plate.

Force on the electron due to electric field : $F_{E} = e E = (1.6 \times 10^{- 19} \times 9.1 \times 10^{2}) = 14.56 \times 10^{- 17} N$

Acceleration of the electron in the vertically upward direction: $a = \frac{F_{E}}{m} = \frac{14.56 \times 10^{- 17} N}{9.1 \times 10^{- 31} kg} = 1.6 \times 10^{14} m / s^{2}$

No net force is acting in the horizontal direction on the electron. Hence, the velocity in the horizontal direction will not change and it will remain equal to $2 \times 10^{6} m s^{- 1}$ . But will remain accelerated in vertically upward direction. Let after covering a horizontal distance equal to $x$ , the electron strikes the upper plate after the time $t$ . For horizontal motion of electron:

$x = \frac{v_{x}}{t}$

$\Rightarrow t = \frac{x}{v_{x}} = \frac{x}{2 \times 10^{6}} s$

As the electron has to hit the upper plate after the time $t$ , the vertical distance covered $(S_{v})$ after this time will be $0.5 \times 10^{- 2} m$ . Using the second equation of motion:

$(|e| = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg)$

$\Rightarrow x^{2} = \frac{0.5 \times 10^{- 2} \times 8 \times 10^{12}}{1.6 \times 10^{14}} = 25 \times 10^{- 5} m^{2}$

$\Rightarrow x = \frac{5}{\sqrt{10}} \times 10^{- 2} m$

$\Rightarrow x = \frac{5}{\sqrt{10}} cm = 1.58 cm = 1.6 cm$

Hence, the electron will hit the upper plate after covering a distance equal to $1.6 cm$ .

Important Questions on Electric Charges and Fields

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