Question

Two points p and q are maintained at the potential of 10v and -4v respectively. Work done in moving 100 electrons from p to q is

Answer 1

W=V*Q

W=(10-(-4))*(1.6*10^-19)*100

=22.4*10^-17

Answer 2

$Given \\ </p><p> </p><p> </p><p></p><p>✭ \displaystyle\sf n = 100n=100</p><p>$

$✭ \displaystyle\sf V_1 = 10 \ V \ (Initial)V </p><p>1</p><p> </p><p> =10 V (Initial)</p><p>$

$✭ \displaystyle\sf V_2 = -4 \ V \ (Final)V </p><p>2</p><p> </p><p> =−4 V (Final)</p><p>$

$\displaystyle\large\underline{\sf\blue{To \ Find}} </h2><h2></h2><h2>$

$◈ The \: work \: done? \\ \\ </p><p></p><p>\displaystyle\large\underline{\sf\gray{Solution}} </p><p>$

So here we shall use the below mentioned formulas,

$\displaystyle\sf \underline{\boxed{\sf q = ne}} </p><p>q=ne</p><p>$

$</p><p>\displaystyle\sf\underline{\boxed{\sf V = \dfrac{W}{q}}} </p><p>V= </p><p>q</p><p>W</p><p> </p><p>$

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

:

Considering e to be

$1.6 × 10‾¹⁹$

$➝ \displaystyle\sf q = neq=ne</p><p></p><p> \\ ➝ \displaystyle\sf q = 100\times 1.6\times 10^{-19} \\ q=100×1.6×10 </p><p>−19$

$➝ \displaystyle\sf \purple{q = 1.6\times 10^{-17} \ C}$

So now,

$➳ \displaystyle\sf V = \dfrac{W}{q}V= </p><p>q</p><p>W</p><p> </p><p>$

$➳ \displaystyle\sf V_{in} - V_{fi} = \dfrac{w}{q}$

$➳ \displaystyle\sf 10-(-4) = \dfrac{w}{1.6\times 10^{-17}}10−(−4)= </p><p>1.6×10 </p><p>−17</p><p> </p><p>w</p><p>$

$➳ \displaystyle\sf 14 = \dfrac{w}{1.6\times 10^{-17}} \\ 14= </p><p>1.6×10 </p><p>−17</p><p> </p><p>w</p><p>$

$➳ \displaystyle\sf 14\times 1.6\times 10^{-17} = w14×1.6×10 </p><p>−17</p><p> =w$

$➳ \displaystyle\sf \pink{w = 22.4\times 10^{-17} \ J}$

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