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Home » Suppose that the particle in Exercise is an electron projected with velocity . If E between the plates separated by is , where will the electron strike the upper plate?
Suppose that the particle in Exercise 1.33 is an electron projected with velocity \mathrm{v}_{\mathrm{x}}=\mathbf{2 . 0} \times 10^{6} \mathrm{~m} \mathrm{s}^{-1}. If E between the plates separated by 0.5 \mathrm{~cm} is 9.1 \times 10^{2} \mathrm{~N} / \mathrm{C}, where will the electron strike the upper plate? \left(|\mathrm{e}|=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}_{-}\right)
CBSE | Class 12 | Electric Charges and Fields | NCERT | Physics
Suppose that the particle in Exercise 1.33 is an electron projected with velocity \mathrm{v}_{\mathrm{x}}=\mathbf{2 . 0} \times 10^{6} \mathrm{~m} \mathrm{s}^{-1}. If E between the plates separated by 0.5 \mathrm{~cm} is 9.1 \times 10^{2} \mathrm{~N} / \mathrm{C}, where will the electron strike the upper plate? \left(|\mathrm{e}|=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}_{-}\right)

Solution:

Given: Velocity of the electron, \mathrm{v}_{\mathrm{x}}=2.0 \times 10^{6} \mathrm{~m} \mathrm{~s}^{-1}

Separation between the plates, d=0.5 \mathrm{~cm}=0.5 \times 10^{-2} \mathrm{~m}

Electric field between the plates =9.1 \times 10^{2} \mathrm{~N} / \mathrm{C}

We know that, s=\frac{1}{2}\left(\frac{q E}{m}\right)\left(\frac{L}{v_{x}}\right)^{2}

So derriving from the above equation we have, the distance at which the electron strike the upper plate is

L=\sqrt{\frac{2 d m v_{x}^{2}}{q E}}

Substituting the value of given parameters we have,

L=\sqrt{\frac{2 \times 0.005 \times 9.1 \times 10^{-31} \times\left(2 \times 10^{6}\right)^{2}}{1.6 \times 10^{-19} \times 9.1 \times 10^{2}}}

=\sqrt{\frac{0.364 \times 10^{-19}}{14.56 \times 10^{-17}}}

=\sqrt{2.5 \times 10^{-4}}

=1.6 \mathrm{~cm}

The electron strikes the upper plate after 1.6 cm.

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