LEARNATHON
III

Competition for grade 6 to 10 students! Learn, solve tests and earn prizes!

### Theory:

Gravity plays no part in the movement of charges in a conducting metallic wire; electrons travel only when there is a difference in electric pressure – known as the potential difference – along the conductor. A battery consisting of one or more cells can create this potential difference.

Even in the absence of current, internal chemical action causes a potential difference between the terminals of the cell. When a cell is connected to an electric circuit, the potential difference makes the charges to move, thus producing current. The cell has to spend all of its chemical energy to maintain the current flow in a circuit.

Some amount of work is done to bring a positive charge from infinity. But, the work done to bring a positive charge from one point to another within an electric field is known as the potential difference between the two points.
The electric potential difference between two points is defined as the amount of work done in moving a unit positive charge from one point to another point against the electric force.
The potential difference between two points

If charge $$Q$$ is moved from point $$A$$ to point $$B$$, then $$W$$ is the work done to move the charge from point ($$A$$ to point $$B$$).

The formula is given as,

Where, $$V$$ is the potential difference, $$W$$ is the work done, and $$Q$$ is the charge in coulomb ($$C$$).

The difference in the electric potential of the two points gives the potential difference.

Consider $$V_{A}$$ as the electric potential at point $$A$$ and $$V_{B}$$ as the electric potential at point $$B$$. Now, the potential difference between the two points $$A$$ and $$B$$ is written as,

1. When $$V_{A}$$ is more than $$V_{B}$$.

2. When $$V_{B}$$ is more than $$V_{A}$$.

Unit:
The SI unit of potential difference is volt ($$V$$). The unit 'volt' is named after Alessandro Volt, an Italian physicist.
The potential difference between two points is one volt if one joule of work is done in moving one coulomb of charge from one point to another against the electric force.
$\begin{array}{l}1\phantom{\rule{0.147em}{0ex}}\mathit{volt}\phantom{\rule{0.147em}{0ex}}=\frac{1\phantom{\rule{0.147em}{0ex}}\mathit{joule}}{1\phantom{\rule{0.147em}{0ex}}\mathit{coulomb}}\\ \\ 1\phantom{\rule{0.147em}{0ex}}V\phantom{\rule{0.147em}{0ex}}=1\phantom{\rule{0.147em}{0ex}}J{C}^{-1}\end{array}$