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Electric Field - Intensity of Electric Field due to a Continuous Charge Distribution

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Estimated time: 10 minutes
CISCE: Class 12

Introduction

Coulomb's Law and the Principle of Superposition work well for a small, countable number of point charges. However, real objects — rods, rings, discs, spheres — contain an enormous number of charges spread continuously over their length, surface, or volume.

Instead of adding forces from billions of individual charges, physicists treat the charge as smeared continuously over space and replace summation with integration. This is the core idea behind continuous charge distributions.

Analogy: Think of a charged rod as a line made of infinitely many "grains of charge-sand." Each grain is too small to matter alone, but adding up (integrating) their tiny contributions gives the total electric field — just like adding countless grains of sand gives a measurable heap.

CISCE: Class 12

Key Terms

Term Definition SI Unit
Linear charge density (λ) Charge per unit length C/m
Surface charge density (σ) Charge per unit area C/m2
Volume charge density (ρ) Charge per unit volume C/m3
dq Infinitesimally small element of charge C
\[\vec r_1\] Position vector of the charge element m
\[\vec r_2\]​ Position vector of the field point PP m
\[r_{21}\] Distance between charge element and field point m
\[\hat r_{21}\] Unit vector from charge element to field point dimensionless
CISCE: Class 12

Types of Continuous Charge Distribution

Linear Charge Distribution

Charge is distributed along a length (e.g., a charged wire or rod).

dq = λ dl
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\int_L\frac{\lambda dl}{r_{21}^2}\hat{r}_{21}\]

Example: A uniformly charged straight wire or a charged circular ring.

Surface Charge Distribution

Charge is distributed over a two-dimensional surface (e.g., a charged plate or disc).

dq = σ dS
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\int_S\frac{\sigma dS}{r_{21}^2}\hat{r}_{21}\]

Example: A uniformly charged conducting plate or charged disc.

Volume Charge Distribution

Charge is distributed throughout a three-dimensional region (e.g., a charged sphere).

dq = ρ dV
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\int_V\frac{\rho dV}{r_{21}^2}\hat{r}_{21}\]

Example: A uniformly charged insulating sphere.

CISCE: Class 12

Derivation Logic

  • Consider a small charge element dq at position \[\vec r_1\]​.
  • By Coulomb's Law, the field due to this element at field point P (position \[\vec r_2\]) is:
    \[d\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{dq}{r_{21}^2}\hat{r}_{21}\]
  • Replace dq with λ dl, σ dS, or ρ dV depending on the geometry.
  • Integrate over the entire length, surface, or volume to account for contributions from every element.
  • The resultant vector integral gives the net electric field at P.
CISCE: Class 12

Key Points: Intensity of Electric Field due to a Continuous Charge Distribution

  • Continuous charge distributions require integration, not simple addition.
  • Three types exist: linear (λ), surface (σ), and volume (ρ).
  • The general method: apply Coulomb's Law to an element dq, then integrate over the full body.
  • \[\hat r_{21}\] varies across the body — always check symmetry before integrating.
  • Gauss's Law is the preferred shortcut for symmetric distributions.
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