Charge Density Formula

Charge Density Formulas | Solved Example Questions

The charge density is a measure of how much electric charge is accumulated in a particular field. Specifically, it finds the charge density per unit volume, surface area, and length.
It measures the amount of electric charge:
(i) per unit length (linear charge density),
(ii) per unit area (surface charge density),
(iii) per unit volume (volume charge density)
Charge density depends on distribution of charge and it can be positive or negative.

Depending on the nature, charge density formula can be given by,
(i) Linear charge density; \[\lambda = \frac{q}{l}\], where q is the charge and \[l\]is the length over which it is distributed. The SI unit is Cm–1.
(ii) Surface charge density; \[\sigma = \frac{q}{A}\], where, q is the charge and A is the area of the surface. The SI unit is Cm–2.
(iii) Volume charge density; \[\rho = \frac{q}{V}\], where q is the charge and V is the volume of distribution. The SI unit is Cm–3. 

Example: A long thin rod of length 50 cm has a total charge of 5 mC uniformly distributed over it. Find the linear charge density.
q = 5 mC = 5 × 10–3 C, \[l\]= 50 cm = 0.5 m. \[\lambda \]= ?
\[\lambda = \frac{q}{l} = \frac{{5 \times {{10}^{--3}}}}{{0.5}} = {10^{--2}}C{m^{--1}}\]

Example: A cuboidal box penetrates a huge plane sheet of charge with uniform surface charge density 2.5×10–2 Cm–2 such that its smallest surfaces are parallel to the sheet of charge. If the dimensions of the box are 10 cm × 5 cm × 3 cm, then find the charge enclosed by the box.
Charge enclosed by the box = charge on the portion of the sheet enclosed by the box.
The area of the sheet enclosed; A = area of the smallest surface of the box
= 5 cm × 3 cm = 15 cm2 = 15 × 10–4 m2
Charge density; \[\sigma \]= 2.5 ×10–2 Cm–2
Charge enclosed; \[q = \sigma A = 2.5 \times {10^{--2}} \times 15 \times {10^{--4}} = 37.5 \times {10^{--6}}C = 37.5\,\mu C\]

A cubical distribution of charge has a uniform volume charge density of \[6\sqrt 3 \times {10^{--2}}C{m^{--3}}\]. The total charge in the distribution is 20 mC. If the same charge is uniformly distributed over the entire volume of a sphere of diameter equal to the leading diagonal of the cube, then the new volume charge density will be:
(a) \[6\sqrt 3 \times {10^{--2}}C{m^{--3}}\]
(b) 3.8 ×10–2 Cm–3
(c) 6 ×10–2 Cm–3
(d) 2 ×10–2 Cm–3
Answer: (b)