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Electric Charges and Fields - Continuous Charge Distribution

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Continuous Charge Distribution refers to a system in which the charge is distributed continuously over a region (line, surface, or volume) rather than being concentrated at discrete points.

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Linear Charge Density (λ\lambda): When the charge is distributed along a line (like a thin wire), λ\lambda is defined as the charge per unit length, given by λ=dqdl\lambda = \frac{dq}{dl}. Its SI unit is C⋅m−1C \cdot m^{-1}.

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Surface Charge Density (σ\sigma): When the charge is distributed over a surface (like a thin sheet), σ\sigma is defined as the charge per unit area, given by σ=dqdS\sigma = \frac{dq}{dS}. Its SI unit is C⋅m−2C \cdot m^{-2}.

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Volume Charge Density (ρ\rho): When the charge is distributed over a three-dimensional volume (like a sphere), ρ\rho is defined as the charge per unit volume, given by ρ=dqdV\rho = \frac{dq}{dV}. Its SI unit is C⋅m−3C \cdot m^{-3}.

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The total electric field at a point due to a continuous charge distribution is obtained by integrating the fields due to infinitesimal charge elements dqdq: E⃗=14πϵ0∫dqr2r^\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{dq}{r^2} \hat{r}.

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Electric Field due to an Infinitely Long Straight Charged Wire: The magnitude of the electric field at a distance rr from the wire is E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}.

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Electric Field due to a Uniformly Charged Infinite Plane Sheet: The electric field is uniform and independent of the distance from the sheet, given by E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}.

📐Formulae

λ=dqdl\lambda = \frac{dq}{dl}

σ=dqdS\sigma = \frac{dq}{dS}

ρ=dqdV\rho = \frac{dq}{dV}

E⃗=14πϵ0∫λdlr2r^\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{\lambda dl}{r^2} \hat{r}

E⃗=14πϵ0∫σdSr2r^\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{\sigma dS}{r^2} \hat{r}

E⃗=14πϵ0∫ρdVr2r^\vec{E} = \frac{1}{4\pi\epsilon_0} \int \frac{\rho dV}{r^2} \hat{r}

E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}

E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}

💡Examples

Problem 1:

A thin wire of length L=0.5 mL = 0.5\text{ m} carries a uniform linear charge density λ=2×10−6 C/m\lambda = 2 \times 10^{-6}\text{ C/m}. Calculate the total charge QQ on the wire.

Solution:

Q=λ×LQ = \lambda \times L Q=(2×10−6 C/m)×(0.5 m)Q = (2 \times 10^{-6}\text{ C/m}) \times (0.5\text{ m}) Q=1×10−6 CQ = 1 \times 10^{-6}\text{ C}

Explanation:

To find the total charge when the density is uniform, we multiply the linear charge density λ\lambda by the total length LL of the wire.

Problem 2:

An infinite plane sheet of charge has a surface charge density σ=8.85×10−12 C/m2\sigma = 8.85 \times 10^{-12}\text{ C/m}^2. Find the magnitude of the electric field EE near the sheet. (Take ϵ0=8.85×10−12 C2N−1m−2\epsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2})

Solution:

E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0} E=8.85×10−122×(8.85×10−12)E = \frac{8.85 \times 10^{-12}}{2 \times (8.85 \times 10^{-12})} E=12=0.5 N/CE = \frac{1}{2} = 0.5\text{ N/C}

Explanation:

The electric field due to an infinite plane sheet is given by E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}. Substituting the given values leads to a constant field of 0.5 N/C0.5\text{ N/C}.