Applications-Of-Gausss-Law L-4 Electrostatics
→ \rightarrow → → \rightarrow → Electrostatics → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Gauss’s Law
Φ = q 1 + q 2 ϵ 0 \begin{aligned}\Phi=\frac{q_1+q_2}{\epsilon_0}\end{aligned} Φ = ϵ 0 q 1 + q 2
Φ = ∑ q i ϵ 0 \Phi=\frac{\sum q_i}{\epsilon_0} Φ = ϵ 0 ∑ q i
Flux: E ⃗ ⋅ d s ⃗ \vec{E} \cdot d \vec{s} E ⋅ d s
→ \rightarrow → → \rightarrow → Gauss's Law → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4
Φ ˉ = ∮ E ⃗ ⋅ d A ⃗ = Q enc ϵ 0 \bar{\Phi}=\oint \vec{E} \cdot d \vec{A}=\frac{Q_{\text {enc }}}{\epsilon_0} Φ ˉ = ∮ E ⋅ d A = ϵ 0 Q enc
Electrostatics → \rightarrow → → \rightarrow → Integral Form → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Electric Field
Excess charge : Q
at the surface of the conductor
Gauss's Law → \rightarrow → → \rightarrow → Electric Field → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Electric Field due to Sphere
Field produce by a charged conducting sphere
σ = Q 4 π R 2 \sigma=\frac{Q}{4 \pi R^2} σ = 4 π R 2 Q
∣ E ⃗ ∣ |\vec{E}| ∣ E ∣ E ⃗ \vec{E} E r ^ \hat r r ^
Integral Form → \rightarrow → → \rightarrow → Electric Field due to Sphere → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Electric Field due to Sphere
∮ E ⃗ ⋅ d A ⃗ = Q enc ϵ 0 \oint \vec{E} \cdot d \vec{A}=\frac{Q_{\text {enc }}}{\epsilon_0} ∮ E ⋅ d A = ϵ 0 Q enc
∮ E d A = Q enc ϵ 0 \oint {E} \ d {A}=\frac{Q_{\text {enc }}}{\epsilon_0} ∮ E d A = ϵ 0 Q enc
E ∮ d A = Q enc ϵ 0 {E} \oint \ d {A}=\frac{Q_{\text {enc }}}{\epsilon_0} E ∮ d A = ϵ 0 Q enc
E. 4 π r 2 = Q enc ϵ 0 = Q ϵ 0 4 \pi r^2=\frac{Q_\text { enc }}{\epsilon_0} = \frac{Q}{\epsilon_0} 4 π r 2 = ϵ 0 Q enc = ϵ 0 Q
Electric Field → \rightarrow → → \rightarrow → Electric Field due to Sphere → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Electric Field due to Sphere
E = Q 4 π ϵ 0 r 2 E=\frac{Q}{4 \pi \epsilon_0 r^2} E = 4 π ϵ 0 r 2 Q
E ⃗ = Q 4 π ϵ 0 r 2 r ^ \vec{E}=\frac{Q}{4 \pi \epsilon_0 r^2} \hat{r} E = 4 π ϵ 0 r 2 Q r ^
σ = Q 4 π R 2 \sigma=\frac{Q}{4 \pi R^2} σ = 4 π R 2 Q
r=R
E ⃗ = Q 4 π ϵ 0 R 2 r ^ = σ ϵ 0 r ^ \vec{E}=\frac{Q}{4 \pi \epsilon_0 R^2} \hat{r} =\frac{\sigma}{\epsilon_0} \hat{r} E = 4 π ϵ 0 R 2 Q r ^ = ϵ 0 σ r ^
= σ ϵ 0 n ^ =\frac{\sigma}{\epsilon_0} \hat{n} = ϵ 0 σ n ^
Electric Field due to Sphere → \rightarrow → → \rightarrow → Electric Field due to Sphere → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Field due to a Line Charge
Field due to a line charge density(infinitely long)
Charge per unit length:
d E = λ d z 4 π ϵ 0 s 2 cos θ d E=\frac{\lambda d z}{4 \pi \epsilon_0 s^2} \cos \theta d E = 4 π ϵ 0 s 2 λ d z cos θ
s 2 = r 2 + z 2 s^2=r^2+z^2 s 2 = r 2 + z 2
dE= λ d z 4 π ϵ 0 ( r 2 + z 2 ) r r 2 + z 2 \frac{\lambda d z}{4 \pi \epsilon_0\left(r^2+z^2\right)} \frac{r}{\sqrt{r^2+z^2}} 4 π ϵ 0 ( r 2 + z 2 ) λ d z r 2 + z 2 r
Electric Field due to Sphere → \rightarrow → → \rightarrow → Field due to a Line Charge → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Field due to a Line Charge
d E = λ d z r 4 π ϵ 0 ( r 2 + z 2 ) 3 / 2 d E=\frac{\lambda dzr}{4 \pi \epsilon_0\left(r^2+z^2\right)^{3 / 2}} d E = 4 π ϵ 0 ( r 2 + z 2 ) 3/2 λ d zr
E = λ r 4 π ϵ 0 ∫ − ∞ + ∞ d z ( r 2 + z 2 ) 3 / 2 E=\frac{\lambda r}{4 \pi \epsilon_0} \int_{-\infty}^{+\infty} \frac{d z}{\left(r^2+z^2\right)^{3 / 2}} E = 4 π ϵ 0 λ r ∫ − ∞ + ∞ ( r 2 + z 2 ) 3/2 d z
z = r tan ϕ z=r \tan \phi z = r tan ϕ
d z = r sec 2 ϕ d ϕ d z=r \sec ^2 \phi d \phi d z = r sec 2 ϕ d ϕ
r 2 + z 2 = r 2 + r 2 tan 2 ϕ = r 2 sec 2 ϕ r^2+z^2=r^2+r^2 \tan ^2 \phi=r^2 \operatorname{sec}^2 \phi r 2 + z 2 = r 2 + r 2 tan 2 ϕ = r 2 sec 2 ϕ
Electric Field due to Sphere → \rightarrow → → \rightarrow → Field due to a Line Charge → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Field due to a Line Charge
E = λ r 4 π ϵ 0 ∫ − π / 2 + π / 2 r sec 2 ϕ d ϕ r 3 sec 3 ϕ E =\frac{\lambda r}{4 \pi \epsilon_0} \int_{-\pi / 2}^{+\pi / 2} \frac{r \sec ^2 \phi d \phi}{r^3 \sec ^3 \phi} E = 4 π ϵ 0 λ r ∫ − π /2 + π /2 r 3 s e c 3 ϕ r s e c 2 ϕ d ϕ
= λ 4 π ϵ 0 r ∫ − π / 2 π / 2 cos ϕ d ϕ =\frac{\lambda}{4 \pi \epsilon_0 r} \int_{-\pi / 2}^{\pi / 2} \cos \phi d \phi = 4 π ϵ 0 r λ ∫ − π /2 π /2 cos ϕ d ϕ
= λ 4 π ϵ 0 r sin ∣ − π / 2 π / 2 =\frac{\lambda}{4 \pi \epsilon_0 r} \sin |_{-\pi / 2}^{\pi / 2} = 4 π ϵ 0 r λ sin ∣ − π /2 π /2
E = λ 2 π ϵ 0 r E =\frac{\lambda}{2 \pi \epsilon_0 r} E = 2 π ϵ 0 r λ
Field due to a Line Charge → \rightarrow → → \rightarrow → Field due to a Line Charge → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Field due to a Line Charge
E ⃗ = λ r ^ 2 π ϵ r \vec{E}=\frac{\lambda \hat{r}}{2 \pi \epsilon_r} E = 2 π ϵ r λ r ^
Field due to a Line Charge → \rightarrow → → \rightarrow → Field due to a Line Charge → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Gaussian Surface
E ⋅ 2 π r ⋅ l = Q enc ϵ 0 = λ l ϵ 0 \cdot 2 \pi r \cdot l=\frac{Q_\text{enc}}{\epsilon_0}=\frac{\lambda l}{\epsilon_0} ⋅ 2 π r ⋅ l = ϵ 0 Q enc = ϵ 0 λ l
Field due to a Line Charge → \rightarrow → → \rightarrow → Gaussian Surface → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Gaussian Surface
E ⃗ = λ 2 π ϵ 0 r r ^ \vec{E}=\frac{\lambda}{2 \pi \epsilon_0 r} \hat{r} E = 2 π ϵ 0 r λ r ^
Field due to a Line Charge → \rightarrow → → \rightarrow → Gaussian Surface → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Field due to a Infinite Sheet
Field due to an infinite sheet of surfaces charge σ \sigma σ
Gaussian Surface → \rightarrow → → \rightarrow → Field due to a Infinite Sheet → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Gaussian Surface
Total electric flux = E.2A
Total charge enclosed = σ \sigma σ
Gaussian Surface → \rightarrow → → \rightarrow → Gaussian Surface → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Gaussian Surface
Field due to a Infinite Sheet → \rightarrow → → \rightarrow → Gaussian Surface → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Thin Conducting Plate
Gaussian Surface → \rightarrow → → \rightarrow → Thin Conducting Plate → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Gaussian Law
∮ E ⃗ ⋅ d A ⃗ = Q enc ϵ 0 \oint \vec{E} \cdot d \vec{A}=\frac{Q_{\text {enc }}}{\epsilon_0} ∮ E ⋅ d A = ϵ 0 Q enc
Gaussian Surfac → \rightarrow → → \rightarrow → Gaussian Law → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Problem
Positive charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R. Using Gauss’s law obtain the electric field inside and outside the sphere.
Thin Conducting Plate → \rightarrow → → \rightarrow → Problem → \rightarrow → → \rightarrow →
Applications-Of-Gausss-Law L-4 Thank You
Gaussian Law → \rightarrow → → \rightarrow → Thankyou → \rightarrow → → \rightarrow →
Resume presentation
Applications-Of-Gausss-Law L-4 Electrostatics $\rightarrow$ $\rightarrow$ Electrostatics $\rightarrow$ Gauss's Law $\rightarrow$ Integral Form
