JEE Main 2025 Jan 22 Shift 1, Physics Q9: Coulomb's Law And Electric Field

JEE Main2025Jan 22, Shift 1Physics

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 field region with a horizontal component of velocity $1 0^{6}$ m/s. If the magnitude of the electric field intensity between the plates is 9.1 V/cm, then the vertical component of velocity of electron is

(mass of electron $= 9.1 \times 1 0^{- 31}$ kg and charge of electron $= 1.6 \times 1 0^{- 19}$ C)

A
$1 \times 1 0^{6}$ m/s
B
0
C
$16 \times 1 0^{6}$ m/s
D
$16 \times 1 0^{4}$ m/s

Solution

Time inside the field: $t = \frac{L}{v _{x}} = \frac{0.10}{1 0 ^{6}} = 1 0^{- 7}$ s.

Field $E = 9.1$ V/cm $= 910$ V/m.

Vertical acceleration: $a = \frac{e E}{m} = \frac{1.6 \times 1 0 ^{- 19} \times 910}{9.1 \times 1 0 ^{- 31}} = 1.6 \times 1 0^{14}$ m/s $^{2}$ .

Vertical velocity gained: $v_{y} = a t = 1.6 \times 1 0^{14} \times 1 0^{- 7} = 1.6 \times 1 0^{7} = 16 \times 1 0^{6}$ m/s.

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JEE Main 2025 Jan 22 Shift 1, Physics Q9: Coulomb's Law And Electric Field

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