JEE MAIN - Physics (2025 - 22nd January Morning Shift - No. 15)
An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates, each of 10 cm length. The electron emerges out of the electric field region with a horizontal component of velocity $10^6 \mathrm{~m} / \mathrm{s}$. If the magnitude of the electric field between the plates is $9.1 \mathrm{~V} / \mathrm{cm}$, then the vertical component of velocity of electron is
(mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ and charge of electron $=1.6 \times 10^{-19} \mathrm{C}$ )
$1 \times 10^6 \mathrm{~m} / \mathrm{s}$
$16 \times 10^6 \mathrm{~m} / \mathrm{s}$
$16 \times 10^4 \mathrm{~m} / \mathrm{s}$
0
Explanation
_22nd_January_Morning_Shift_en_15_1.png)
No force in horizontal direction, i.e., $F_x=0$
$\Rightarrow a_x=0$
So, $v_x$ = constant
hence, $t=\frac{l}{v_x}$ ..... (1)
Now, for y-direction, using Ist equation of motion,
$${v_y} = {u_y} + {a_y}t$$
$$ \Rightarrow {v_y} = 0 + {{eE} \over m}\left( {{l \over {{v_x}}}} \right)$$ (from (1) and usiung $F=ma$ and $a=\frac{F}{m}$)
$$ \Rightarrow {v_y} = {{1.6 \times {{10}^{ - 19}} \times 9.1 \times {{10}^2} \times 0.1} \over {9.1 \times {{10}^{ - 31}} \times {{10}^6}}}$$
$$ = 1.6 \times {10^7}$$ m/s
$$ \Rightarrow {v_y} = 16 \times {10^6}$$ m/s.
Comments (0)
