An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2 × 104 N/C. The time taken by the electron to fall this distance is

Question

An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2 × 104 N/C. The time taken by the electron to fall this distance is (A) 1.3 × 102 s (B) 2.1 × 10-12 s (C) 1.6 × 10-10 s (D) 2.9 × 10-9 s

Tardigrade Admin · Accepted Answer

Here, the direction of field is upward.
So the negatively charged electron experiences a downward force.

∴

The acceleration of electron is

a_{e} = \frac{e E}{m _{e}} ... (i)

The time required by the electron to fall through a distance

h

is

t_{e} = (\frac{2 h}{A a _{e}}) = \frac{2 h m _{e}}{e E}

(using (i))

= [\frac{2 \times 1.5 \times 1 0 ^{- 2} \times 9.11 \times 1 0 ^{- 31}}{1.6 \times 1 0 ^{- 19} \times 2 \times 1 0 ^{4}}]^{^{1/2}}

= 2.9 \times 1 0^{- 9} s

Q. An electron initially at rest falls a distance of $1.5 c m$ in a uniform electric field of magnitude $2 \times 1 0^{4} N / C$ . The time taken by the electron to fall this distance is

$1.3 \times 1 0^{2} s$

$2.1 \times 1 0^{- 12} s$

$1.6 \times 1 0^{- 10} s$

$2.9 \times 1 0^{- 9} s$

Q. An electron initially at rest falls a distance of 1.5cm in a uniform electric field of magnitude 2×104N/C. The time taken by the electron to fall this distance is

1.3×102s

2.1×10−12s

1.6×10−10s

2.9×10−9s

Q. An electron initially at rest falls a distance of $1.5 c m$ in a uniform electric field of magnitude $2 \times 1 0^{4} N / C$ . The time taken by the electron to fall this distance is

$1.3 \times 1 0^{2} s$

$2.1 \times 1 0^{- 12} s$

$1.6 \times 1 0^{- 10} s$

$2.9 \times 1 0^{- 9} s$