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Old Jul 1st 2019, 03:02 AM   #1
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How to solve this

Hello sir
How to solve this
The amount of work done in moving a charge of 10C from a point of 180 volts to another point of 210 volt is
1. 1800J
2. 300J
3. 3900J
4. 3000J
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Old Jul 1st 2019, 03:45 AM   #2
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Originally Posted by TYUTNTT View Post
Hello sir
How to solve this
The amount of work done in moving a charge of 10C from a point of 180 volts to another point of 210 volt is
1. 1800J
2. 300J
3. 3900J
4. 3000J
$\displaystyle W = Q \Delta V$

So....

-Dan
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Old Jul 1st 2019, 10:21 PM   #3
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W=qΔv

w=10c*30v
w=300j
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Old Jul 2nd 2019, 06:23 PM   #4
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Originally Posted by TYUTNTT View Post
W=qΔv

w=10c*30v
w=300j
Good! However a quick note: The unit "volts" is represented as "V" and "Joules" is "J". Capitalization is important.

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Old Jul 3rd 2019, 03:11 AM   #5
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Naming Conventions

I have a vague recollection from somewhere that the units that are named after a person use a Capital first letter, and thus a Capital abbreviation
Thus:
V for Volts (named after Alessandro Volta),
J for Joule (James Prescott Joule),
W for Watt (James Watt),
etc.
But lower case for other units.
Thus:
m for metre,
g for gram,
etc.
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