Electrostatic Potential and Capacitance
Last Updated : 07 May, 2025
Electric potential and capacitance originate from the concept of charge. The charge is determined by comparing the number of protons and electrons present in a material. A material with more protons than electrons carries a positive charge, while more electrons than protons carries a negative charge.
If the numbers are equal, the material is neutral. Charges exert electrical forces, with opposite charges attracting and like charges repelling. Charge is measured in coulombs (C), with protons and electrons each having a charge of +1.602 × 10−19 C and −1.602 × 10−19 C, respectively.
Electrostatic Potential
The electrostatic potential (V) at a point in an electrostatic field is the work needed to move a unit positive charge (without acceleration) from infinity to that point.
The unit for electric potential is joules per coulomb (J/C), which measures the amount of work per charge. The J/C unit is commonly called a volt (V).
V = W/q
Where,
V = Electrostatic potential (Volts)
W = Work done (Joules)
q = Charge (Coulombs)
Electrostatic Potential Due to a Point Charge
The potential at 'a' distance 'r' from a charge 'Q' is given by
𝑽 = 𝒌𝑸/𝑟
Where,
k = Coulomb's constant (9 \times 109 𝑵𝒎2/𝑪2)
Q = Charge (Coulombs)
r = Distance from charge (meters)
Capacitance
A capacitor is a device that stores electrical energy. The capacitance (C) of a capacitor is defined as the ratio of charge stored (Q) to the potential difference (V) applied.
The unit of measurement for capacitance is coulomb per voltage (C/V), representing the amount of charge present per voltage applied. The farad (F) is the standard unit for capacitance commonly used instead of C/V.
𝑪 = 𝑸/𝑽
Where,
C = Capacitance (Farads, F)
Q = Charge stored (Coulombs)
V = Potential difference (Volts)
Capacitance of a Parallel Plate Capacitor
For a capacitor with two parallel plates separated by a distance 'd' and area 'A' the capacitance is given by,
U = \varepsilon 0 𝐀 / D
Where,
\varepsilon0 = Permittivity of free space (8.85 \times 10-12 F/m
A = Plate area (square meters)
d = Separation distance (meters)
Energy Stored in a Capacitor
The energy stored in a charged capacitor is given by,
U = 1/2 CV2
Where,
U = Energy stored (Joules)
C = Capacitance (Farads)
V = Voltage (Volts)
Practice Problems
1. What is the electrostatic potential at a distance of 5 m from a charge of 2 × 10⁻⁶ C? (k = 9 × 10⁹ Nm²/C²)
Solution:
𝑉 = kQ/r = (9 \times 102)(2 \times 10-6)/5 V = 18 \times 103/5 = 3.6 \times 103 V ;Answer: 3600 V
2. If a capacitor has a charge of 10 μC and a potential difference of 5 V, what is its capacitance?
Solution:
C = Q/V = 10 \times 10-6 /5 C = 2 \times 10-6 F; Answer: 2 μF
3. The plates of a parallel plate capacitor have an area of 0.02 m² and are separated by 0.001 m. Find the capacitance if the permittivity of free space is 8.85 × 10⁻¹² F/m.
Solution:
C = \varepsilon0A/d = (8.85 \times 10-12)(0.02)/0.001 C = 1.77 \times 10-13/0.001 = 1.77 \times 10-10 F ; Answer: 177 μF
Conclusion
Electrostatic potential helps measure the energy required to move charges in an electric field, while capacitance describes a capacitor’s ability to store charge. These concepts play a important role in the behavior of electrical circuits. Capacitors, for example, are used in energy storage, filtering signals, and stabilizing voltage in various electronic devices.