Question

Find the number of electrons transferred between two points kept at a potential

difference of 30 V if 50 J of work is done. ​

Answer 1

Explanation:

V=J/C

V=J/C20=40/C

V=J/C20=40/CC=2

V=J/C20=40/CC=21C=6.25*10^6

V=J/C20=40/CC=21C=6.25*10^62C=(6.25*10^6)*2

V=J/C20=40/CC=21C=6.25*10^62C=(6.25*10^6)*22C=12.5*10^6

Answer 2

$\mathtt{\huge{ \fbox{</strong></p><p><strong>Solution</strong><strong> :)}}}$

Given ,

• Potential difference (v) = 30 v
• Work done (w) = 50 joule

We know that ,

Work done to move a unit charge from one point to another point is called potential difference

$\mathtt{ \fbox{Potential \: difference \: (v) = \frac{w}{q} }}$

Thus ,

30 = 50/q

q = 50/30

q = 1.6 coulomb

Hence , the charge is 1.6 C

We know that ,

The charge on a body is an integral multiple of electrons

$\large \mathtt{ \fbox{Charge \: (q) = ne }}$

Thus ,

$\sf \hookrightarrow 1.6 = n \times 1.6 \times {(10)}^{ - 19} \\ \\ \sf \hookrightarrow n = \frac{ \cancel{1.6}}{ \cancel{1.6} \times {(10)}^{ - 19} } \\ \\ \sf \hookrightarrow n = {10)}^{19}$

Hence , the number of electrons is (10)^19

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