# How to solve this

#### TYUTNTT

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

#### topsquark

Forum Staff
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

• 2 people

W=qΔv

w=10c*30v
w=300j

#### topsquark

Forum Staff
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.

-Dan

#### Woody

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.

• 1 person