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 (Vis defined as the amount of work, (W) needed to move a charge, Q.Mathematically, this relation can be expressed as\(V=\frac WQ\).

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, V is written as

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

Given that \(Q=It\), (1) can be expressed in terms of the energy lost, by current, \(I\) through a potential difference, V for a time, t as

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


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:

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