Question

a charged particle in an electric field falls from rest through a distance d in time t.if the charge on the particle is doubled , the time of fall through the same distance will be
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Answer 1

Answer:

ANSWER

Initial velocity u=0

Force on charge=

$f \: = qe \\ \\ f \: = ma \\ \\ qe \: = ma \\ \\ \frac{qe}{m} = a \\ \\ s = ut + \frac{1}{2}a {t}^{2} \\ \\ d = \frac{1}{2} \times \frac{qe}{m} {t1}^{2} \\ \\ d \: = \frac{1}{2} \times \frac{qe}{m} {t2}^{2} \\ \\ \frac{1}{2} \times \frac{qe}{m} {t1}^{2} \times \frac{1}{2} \times \frac{qe}{m} {t2}^{2} \\ \\ \frac{t1}{ \sqrt{2} } = t2$

Answer 2

Answer

t/√2

Explanation:

given, u  = 0. (since the body falls from rest)

F = charge × e

F = q × e

F is also = mass × acceleration

F = m × a

qe =ma

∴ , a = qe/m

now distance (S) = ut + 1/2 at²

                            = 0*t + 1/2 × qe/m ×t²

in the first case time t₁ = 0 + 1/2 × ge/m × t₁²

in case two time t₂ = 0+ 1/2 × ge/m ×t₂²

combine these two equations:

1/2 × ge/m ×t₁² = 1/2 × ge/m ×t₂²

simplifying we get:

t₂ = t₁/√2