**Electric Potential**

In order to move, something needs force. This is also true in electric current. From the definition that electric current is the flow of charges in a conductor, a force cause it to move. This force is called **electric potential.**

**Electric potential ( V) **is defined as the amount of work, (

**) needed to move a charge,**

*W**.Mathematically, this relation can be expressed as\(V=\frac WQ\).*

**Q**As the electric charges move pass through a conductor, they lose energy resulting to a decrease in electric potential. This change in electric potential is called **electric potential difference **or simply **potential difference.**

Here are some equations involving potential difference and its relationship with different electrical quantities.

The electrical energy, ** E** lost or work done by a charge,

*going through a potential difference,*

**Q****is written as**

*V*\(E=QV\quad \quad \quad \quad (1)\).

Given that \(Q=It\), (1) can be expressed in terms of the energy lost, ** E **by current, \(I\) through a potential difference,

*for a time,*

**V****as**

*t***\(E=VIt \quad \quad \quad \quad (2)\).**

**Example 1.**

What is the potential difference of a car battery if it takes \(1.5 \times 10^2\;C\)* *of charge and uses \(1.8 \times 10^3\;J\) of energy to start a car?

Given:

\(E=1.8\times 10^3\;J\\ Q=1.5\times 10^2\;C\)

Solution:

To solve for the potential difference, we derive the equation for potential difference from **(1)**,

\(E=QV\rightarrow V=\frac EQ \).

Thus, the potential difference of the car battery is

\(V=\frac EQ\\ \;\;\,=\frac {1.8 \times 10^3\;J}{1.5 \times 10^2\;C}\\ \quad \\ \;\;\,=12\;V\)

**Example 2.**

How much energy is needed by a rice cooker if it has a charge of 950 C passing through it with a potential difference of 200 V.

Given:

\(Q=950\;C\\ V=200\;V\)

Solution:

\(E=QV\\ \;\;\,=(950\;C)(200\;V)\\ \;\;\,=190,000\;J\;\text{or}\;1.9 \times 10^5\;J\)

**Example 3.**

Solve for the energy stored in a 6.0-V battery that delivers 0.5 A of current for \(3.0\times 10^3 \;s\).

Given:

\(V=6.0\;V\\ I=0.5\;A\\ t=3.0\times10^3\;s\)

Solution:

\(E=VIt\\ \;\;\,=(6.0\;V)(0.5\;A)(3.0\times 10^3\;s)\\ \;\;\,=9,000\;J\; \text{or}\; 9.0\times 10^3\;J\)