Electric potential is somewhat that relates to the potential energy. Also, it is the work that needs to be done to move a unit charge from a reference point to a precise point inside the field with production acceleration. Moreover, over in this topic, we will learn the electric potential, electric potential formula, formula’s derivation, and solved example.

**Electric Potential**

For understanding the electric potential, firstly let’s discuss potential energy. Potential energy refers to the stored energy that an object has inside it or it’s the energy that every object stores inside it.

Furthermore, a charged particle in an electric field has potential energy and because of the electrostatic force that can act on it. Also, it is often useful to be able to describe the potential energy charge per unit at a certain point or position. Moreover, this potential energy per unit charge is the electric potential.

In simple words, we can say that electric potential is the energy per unit that we use to describe potential energy based on the position of the object. Moreover, it resides inside the object in the form of a charge. Most noteworthy, the unit of potential energy is Joules (J) and one joule is equal to 1 \(kg\cdot m^{2}/s^{2}\). Besides, the unit of charge is the coulomb (C), furthermore, the unit of electric potential in volts (v) that is equal to a Joule per coulomb (J/C).

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**Electric Potential Formula**

The formula of electric potential is the product of charge of a particle to the electric potential.

Potential energy = (charge of the particle) (electric potential)

U = q × V

U = qV

**Derivation of the Electric Potential Formula**

U = refers to the potential energy of the object in unit Joules (J)

q = refers to the charge of the point particle in unit coulombs (C)

V = refers to the electric potential in units Volts equal to Joules per coulomb (V = J/C)

**Solved Example on Electric Potential Formula**

**Example 1**

Suppose a point particle has a charge of -8.0 \(\mu C\). Furthermore, it moves from point A, with electric potential \(V_{A}\) = +200 V to point B, with an electric potential \(V_{B}\) = +600 V. Now, compute the change in potential energy as an outcome of this movement?

**Solution:**

We can find the potential energy \(\Delta U\) by using the formula:

\(\Delta U\) = \(U_{B}\) – \(U_{A}\)

\(\Delta U\) = q\(V_{B}\) – q\(V_{A}\)

\(\Delta U\) = (-8.0 \(\mu C\)) (+600.0 V) – (-8.0 \(\mu C\)) (+200.0 V)

Moreover, the charge is given in terms of the micro-Coulombs (\(\mu C\)) : 1.0\(\mu C\) = 1.0 × \(10^{-6}\)C. Also, the charge needs to be changed in terms to the correct units before soling the question or equation:

\(\Delta U\) = (-8.0 \(\times 10^{-6}\)C) (+600 J/C) – (-8.0 \(\times 10^{-6}\)C) (+200 J/C)

\(\Delta U\) = 0.0032 J

So, the change in the potential energy due to the movement of the point particle is 0.0032 J.

**Example 2**

Now, appoint particle has a charge of +6.0 \(\mu C\). Also, it moves from point A with an electric potential of \(V_{A}\) = -100 V to point B. Moreover, during this process, the potential energy changes by + 0.0018 J. So, find the electric potential at point B?

**Solution:**

For this we have to rearrange the formula to find the electric potential

\(\Delta U\) = \(U_{B}\) – \(U_{A}\)

\(\Delta U\) = \(qV_{B}\) – \(aV_{A}\) \(\rightarrow\) \(V_{B}\frac{\Delta U + aV_{A}}{q}\)

Now, the charge is in terms of micro Coulombs (\(\mu C\)) : 1.0\(\mu C\) = 1.0 × \(10^{-6}\)C. Also, the charge needs to be converted in correct unit before solving the question:

\(V_{B}\) = \(\frac{\Delta U + aV_{A}}{q}\)

\(V_{B}\) = \(\frac{(+ 0.0018 J) + (+6.0 \times 10^{-6}C ) (-100 V)}{(+6.0 \times 10^ {-6})}\)

\(V_{B}\) = \(\frac{(+ 0.0018 J)}{(+6.0 \times 10^ {-6})}\) – 100 V

\(V_{B}\) = 300 V – 100 V

\(V_{B}\) = +200 V

So, the electric potential at point B is +200 Volts.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…