Question

10^9 electrons are removed from a body every second. What will be the charge on the body after 10^9 second

Answer 1

GiveN :

• Number of electrons (n) = 10^9 electrons
• Time interval (t) = 10^9 s

To FinD :

• Charge of the body

SolutioN :

Use formula for Quantization of the charges,

$\large \implies {\boxed{\rm{Q = \pm ne}}}$

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Where,

• n = number of electrons = $\rm{10^9 \: e^{-1}}$
• Q = Charge
• e = Charge on electron = $\rm{1.6 \: \times \:10^{-19} \: C}$

Now, Substitute the values

$\implies \rm{Q = ne}$

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$\implies \rm{Q = (10^9) \times (1.6 \times 10^{-19}}$

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$\implies \rm{Q = 1.6 \times 10^{-19 + 9}}$

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$\implies \rm{Q = 1.6 \times 10^{-10}}$

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$\small \underline{\rm{\therefore \: Charge \: of \: body \: is \: 1.6 \: \times \: 10^{-10} \: C}}$

Answer 2

Explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• $\sf {10}^{9} \: electrons \: are \: removed \: from \: a \: body \: every \: second$

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• $\sf The \: charge \: on \: the \: body \: after \: {10}^{9} \: seconds$

${\green{\underline{\underline{\bold{Solution:-}}}}}$

10⁹ electrons are removed from a body every second.

$\sf Charge \: of \: one \: electron = 1.6 \times {10}^{ - 19}$

$\boxed{ \sf \pink{Q = ne}}$

Here,

• n = number of electrons (10⁹)

• e = charge in one electron $1.6 \times {10}^{ - 19}$

$\sf Charge \: of \: {10}^{9} \: electrons$

$\sf Q = 1.6 \times {10}^{ - 19} \times {10}^{9}$

$\sf 1.6 \times {10}^{ - 19 + 9}$

$\sf 1.6 \times {10}^{ - 10}$

$\boxed{ \sf \purple{Charge \: on \: the \: body \: after \: {10}^{9} \: seconds = 1.6 \times {10}^{ - 10} }}$