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Physics · 2026 · 3 marks
CBSE 2026 · Region 1 · Set 3 · Q23
Figure shows a narrow beam of electrons entering with a velocity of 3 × 10^7 m / s, symmetrically through the space between two parallel horizontal…
Figure shows a narrow beam of electrons entering with a velocity of $\displaystyle 3 \times 10^{7} \mathrm{~m} / \mathrm{s}$, symmetrically through the space between two parallel horizontal plates $\displaystyle \mathrm{P}_{1} \mathrm{P}_{1}^{\prime}$ and $\displaystyle \mathrm{P}_{2} \mathrm{P}_{2}^{\prime}$ kept $\displaystyle 2$ cm apart.
If each plate is $\displaystyle 3$ cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end $\displaystyle \mathrm{P}_{2}^{\prime}$.
If each plate is $\displaystyle 3$ cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end $\displaystyle \mathrm{P}_{2}^{\prime}$.Marking-scheme solution
Since $\displaystyle \mathrm{a}=\dfrac{\mathrm{qE}}{\mathrm{m}}=\dfrac{\mathrm{qV}}{\mathrm{mL}}$
$\displaystyle t=\dfrac{x}{u_{x}}$
$\displaystyle y=\dfrac{1}{2}\left(\dfrac{\mathrm{qV}}{\mathrm{mL}}\right)\left(\dfrac{x}{u_{x}}\right)^{2}$
$\displaystyle \mathrm{V}=\dfrac{2 y m L u_{x}{ }^{2}}{e x^{2}}$
$\displaystyle \mathrm{V}=\dfrac{2 \times 1 \times 10^{-2} \times 9.1 \times 10^{-31} \times 2 \times 10^{-2}\left(3 \times 10^{7}\right)^{2}}{1.6 \times 10^{-19} \times\left(3 \times 10^{-2}\right)^{2}}$
$\displaystyle \mathrm{V}=2275$ Volt
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