Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Charge Density Formula

ffImage
Last updated date: 22nd Jul 2024
Total views: 63.3k
Views today: 1.63k
hightlight icon
highlight icon
highlight icon
share icon
copy icon

Introduction

The flow of charge is very important for determining the electric field. They have a collection of electric charges. and thus charge density is very important to calculate for many purposes. The charge density is calculated based on the surface area and the volume of the electric object. The charge density formula is a very interesting and important topic.


The charge density is the measure of the accumulation of electric charge in a given particular field. The following are some of the dimensions in which the charge density is measured:

  • 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 will be Coulomb m-1.

  • Surface Charge Density: \[ \sigma = \frac{q}{A}\] where, q is the charge and A is the area of the surface. The SI unit is Coulomb m-2.

  • Volume Charge Density: \[ \rho = \frac{q}{V}\] where q is the charge and V is the volume of distribution. The SI unit is Coulomb m-3.


Charge density is based on the distribution of electric charge and it can be either positive or negative. The measure of electric charge per unit area of a surface is called the charge density.


Example Problems

1. A long thin rod of length 50 cm has a total charge of 5 mC uniformly distributed over it. Find the linear charge density.

Solution:

q = 5 mC = 5 × 10–3 C, l= 50 cm = 0.5 m. λ= ?

\[\lambda = \frac{q}{l} \]

\[= \frac{(5 \times 10^{-3} )} {0.5} \]

\[= 10^{-2} Cm^{-1} \]


2. 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.

Solution:

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 

\[= 5 cm × 3 cm = 15 Cm^{2} = 15\times 10^{-4} m^{2} \]

Charge density; \[\sigma = 2.5 \times 10^{-2} Cm^{-2} \]

Charge enclosed; q= \[\sigma\] A =2.5 \[\times\] 10−2 \[\times\]15 \[\times\]10−4 

= 37.5 \[\times\]10−6

C = 37.5 μC