Electric Potential due to a Point Charge Derivation

ELECTRIC POTENTIAL

    The electric potential (V) at any point is equal to the amount of work done by the external force in bringing the unit positive charge `(q_0)` from infinity to that point inside the electric field without changing the kinetic energy.

Electric Potential due to a Point Charge Derivation

    Suppose a point charge +q is placed at a point O. We have to obtain the value of electric potential at a point P which is at a distance r apart from point O. For it work done in taking a test charge `q_0` from infinity to point P would be calculated.

Electric Potential due to a Point Charge Derivation

 Work done in bringing test charge `q_0` from B to A.

`W = \int_B^A \vecF.d\vecx`

`W = \int_B^A Fdx cos 180^\circ`

`W = \int_B^A F dx ( - 1)`

`W = - \int_B^A F dx `

    Therefore work done in bringing test charge `q_0` from infinity to point P at r distance.

`W = - \int_{\infty}^r F dx `

`W = - \int_{\infty}^r \frac{K q q_0}{x^2} dx `        (OB = x)

`W = - K q q_0 \int_{\infty}^r \frac{1}{x^2} dx `

`\frac{W}{q_0} = - K q \int_{\infty}^r x^{-2} dx `

`V = - K q [\frac{x^{-2 + 1}}{- 2 + 1}]_{\infty}^r`        `(\because \frac{W}{q_0} = V)`

`V = - K q  [\frac{x^{- 1}}{- 1}]_{\infty}^r`

`V = + K q  [\frac{1}{x}]_{\infty}^r`

`V =  K q [\frac{1}{r} - \frac{1}{\infty}]`

`V =  K q [\frac{1}{r} - 0]`

`V =  K q [\frac{1}{r}]`

`V =  \frac{K q}{r}`

    Thus, electric potential decreases with an increase in r.

    The formula `V =  \frac{K q}{r}` is used in calculating the electric potential due to the electric dipole at `r, \theta` point.

Short Answer Type Questions

1. What is the term used to describe the potential difference between two points in an electric circuit? 

Answer: Electric voltage.

2. What is responsible for driving the flow of electric current in a circuit? 

Answer: Electric voltage.

3. What is measured in units of volts and represents the strength of an electric field in a circuit? 

Answer: Electric voltage.

4. What is required to overcome the resistance in a circuit and maintain a steady flow of current? 

Answer: Electric voltage.

5. What is essential for the operation of various electrical devices and appliances? 

Answer: Electric voltage.

Numerical Questions 

1. Given the values of K = 9 `\times 10^9 \frac{N m^2}{C^2}` , q = 5 C, and r = 2m, calculate the value of V. (Answer: V = 22.5 `\times 10^9` V)

NCERT Chapter 2 Physics class 12

• Electric Potential
• Potential due to a Point Charge
• Electric Potential due to Charged Conducting Sphere with Derivation
• Electric Potential Due to Charged Conducting Sphere
• Electric Potential Due to Charged Non-Conducting Sphere
• Electric Potential due to Dipole at any Point
• Equipotential Surfaces and Properties of Equipotential Surface
• Electric Potential Due to Electric Dipole
• Electric Potential Due to a Group of Charges and Relation between Electric Field and Potential
• Electric Potential Energy of a System of Two-Point Charges
• Define the Electrostatic Potential Energy of a System of Two and Three-Point Charges
• Work in Rotation of Electric Dipole in Electric Field
• Potential Energy of a Dipole in an External Field
• Electrostatics of Conductors
• Dielectrics and polarization