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Physics-

General

Easy

Question

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V subscript 2 end subscript minus V subscript 1 end subscript equals 20$ V. ( $i e$ , plate 2 is at a higher potential). The plates are separated by $d equals 0.1$ m and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? (e=1.6 $cross times 10 to the power of negative 19 end exponent C comma m subscript 0 end subscript equals 9.11 cross times 10 to the power of negative 31 end exponent k g right parenthesis$

$2.65 \times 10^{6} {m s}^{- 1}$
$7.02 \times 10^{12} {m s}^{- 1}$
$1.87 \times 10^{6} {m s}^{- 1}$
$32 \times 10^{- 19} {m s}^{- 1}$

The correct answer is: $2.65 cross times 10 to the power of 6 end exponent m s to the power of negative 1 end exponent$

Since $V subscript 2 end subscript greater than V subscript 1 end subscript comma$ so electric field will point from plate 2 to plate 1.
The electron will experience an electric force, opposite to the direction of electric field, and hence move towards the plate 2.

Use work-energy theorem to find speed of electron when it strikes the plate 2.
$fraction numerator m subscript e end subscript v to the power of 2 end exponent over denominator 2 end fraction minus 0 equals e left parenthesis V subscript 2 end subscript minus V subscript 1 end subscript right parenthesis$
Where $v$ is the required speed.
$therefore blank fraction numerator 9.11 cross times 10 to the power of negative 31 end exponent over denominator 2 end fraction v to the power of 2 end exponent equals 1.6 cross times 10 to the power of negative 19 end exponent cross times 20$
$rightwards double arrow blank v equals blank square root of fraction numerator 1.6 cross times 10 to the power of negative 19 end exponent cross times 40 over denominator 9.11 cross times 10 to the power of negative 31 end exponent end fraction end root equals 2.65 cross times 10 to the power of 6 end exponent m s to the power of negative 1 end exponent$

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