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ElenaW [278]
3 years ago
7

If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

gh it?
Physics
1 answer:
ioda3 years ago
3 0

Given :

Number of operations move through a pocket calculator during a full day's operation , n=1.34 \times 10^{20} .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in  one electron is :

e^-=-1.6\times 10^{-19}\ coulombs

So , charge on n electron is :

C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

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alexdok [17]
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3 years ago
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6 0
4 years ago
A constant electric field of magnitude E = 148 V/m points in the positive x-direction. How much work (in J) does it take to move
DochEvi [55]

Answer:

W=-2.1405\times 10^9\,J

Explanation:

Given:

electric field, E=148\,V.m^{-1}

charge, Q=-13\,\mu C=-13\times 10^{-6}\,C

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final position coordinates,   p2 =(107,76)

We find the distance through which the charge has been moved:

d=\sqrt{(x1-x2)^2+(y1-y2)^2}

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d=\sqrt{(107-(-81))^2+(76-(-131))^2}

d= 279.63\,m

Now we need the angle through which displacement is made with respect to the direction of electric field.

tan\,\theta= \frac{y2-y1}{x2-x1}

\theta= tan^{-1}[\frac{76-(-131)}{107-(-81)} ]

\theta= 47.75^{\circ}

Now from the relation between the change in potential difference:

\Delta V= E.d.cos\,\theta

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\Delta V= 27826.06 V

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∴\Delta V=\frac{W}{Q}

W=\frac{\Delta V}{Q}

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W=\frac{27826.06 }{-13\times 10^{-6}}

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3 years ago
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IgorLugansk [536]
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