Application of Gauss' law

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   i. Consider a sphere of radius R with its centre at O, charged to a uniform surface charge density σ placed in a dielectric medium of permittivity ε (ε = ε₀k). The total charge on the sphere, $q=\sigma\times4\pi R^2$. ii. To find the electric field intensity at a point P, at a distance r from the centre of the charged sphere, imagine a concentric Gaussian sphere of radius r passing through P as shown in the figure below. Let ds be a small area around the point P on the Gaussian surface. iii. By Gauss' theorem, the net flux through a closed Gaussian surface, $$\phi=\dfrac{q}{\varepsilon_0}\ \text{(for air/vacuum }k=1\text{)}\qquad....(1)$$ where q is the total charge inside the closed surface. iv. Due to symmetry and spheres being concentric, the electric field at each point on the Gaussian surface has the same magnitude E and it is directed radially outward. Also, the angle between the direction of E and the normal to the surface of the sphere (ds) is zero i.e. $\cos\theta=1$ $$\therefore \vec{E}\cdot\vec{ds}=E\,ds\cos\theta=E\,ds$$ $$\therefore \text{Flux }d\phi\text{ through the area }ds,\ d\phi=E\,ds$$ Total electric flux through the Gaussian surface $$\phi=\oint \vec{E}\cdot\vec{ds}=\oint E\,ds=E\oint ds$$ $$\therefore \phi=E\times4\pi r^2\qquad....(2)$$ v. From equations (1) and (2), $$\dfrac{q}{\varepsilon_0}=E\times4\pi r^2$$ $$\therefore E=\dfrac{q}{4\pi\varepsilon_0 r^2}\qquad....(3)$$ This is the expression for the electric intensity at the point P, outside the charged sphere.

2. 2.↑ back to the question

   i. Consider a uniformly charged infinite plane sheet with uniform surface charge density (charge per unit area) σ kept in a medium of permittivity ε (ε = ε₀k). ii. To find the electric field due to a charged infinite plane sheet at P at a distance r from the sheet, imagine a Gaussian surface around P in the form of a cylinder having cross sectional area A and length 2r with its axis perpendicular to the plane sheet. The plane sheet passes through the middle of the length of the cylinder such that the ends of the cylinder (called end caps P and P') are equidistant (at a distance r) from the plane sheet as shown in the figure below. iii. By Gauss' theorem, the net flux through a closed surface $$\phi=\dfrac{q}{\varepsilon_0}\ \text{(for air/vacuum }k=1\text{)}\qquad....(1)$$ where q is the total charge inside the closed surface. iv. By symmetry, electric field is perpendicular to the plane sheet and directed outwards, having the same magnitude at a given distance on either side of the sheet. The electric field is at right angles to the end caps and away from the plane. Its magnitude is the same at P and P'. The flux passing through the curved surface is zero as the electric field is tangential to this surface. $\therefore$ The total flux through the closed Gaussian surface is given by, $$\phi=\left[\oint \vec{E}\cdot\vec{ds}\right]_P+\left[\oint \vec{E}\cdot\vec{ds}\right]_{P'}+\left[\oint \vec{E}\cdot\vec{ds}\right]_{\text{curved surface}}$$ $$=EA+EA\qquad\left(\because\theta=0\text{ for ends }P\text{ and }P';\ \cos\theta=1\text{ and }\theta=90^\circ\text{ for curved surface};\ \cos\theta=0\right)$$ $$\therefore \phi=2EA\qquad....(2)$$ v. From equations (1) and (2), $$\dfrac{q}{\varepsilon_0}=2EA$$ $$\therefore \dfrac{\sigma A}{\varepsilon_0}=2EA\qquad\left(\because\sigma=\dfrac{q}{A}\right)$$ $$\therefore E=\dfrac{\sigma}{2\varepsilon_0}$$ This is the required expression. vi. Direction of electric field is outward if sheet is positively charged and inward if it is negatively charged.

3. 3.↑ back to the question

   i. Gauss' law: The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε₀. $$\oint \vec{E}\cdot\vec{ds}=\dfrac{q}{\varepsilon_0}$$ where q is the total charge within the surface. ii. Expression for electric field intensity at a point outside an infinitely long charged conducting cylinder: $$E=\dfrac{\lambda}{2\pi k\varepsilon_0 r}$$

4. 4.↑ back to the question

   [From 8.2, Q3 (ii)]

   i. Gauss' law: The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by ε₀. $$\oint \vec{E}\cdot\vec{ds}=\dfrac{q}{\varepsilon_0}$$ where q is the total charge within the surface. ii. Expression for electric field intensity at a point outside an infinitely long charged conducting cylinder: $$E=\dfrac{\lambda}{2\pi k\varepsilon_0 r}$$

5. 5.↑ back to the question

   i. Consider a uniformly charged wire of infinite length having a constant linear charge density λ (charge per unit length), kept in a medium of permittivity ε (ε = ε₀k). ii. To find the electric field intensity at P, at a distance r from the axis of the charged wire, imagine a coaxial Gaussian cylinder of length $l$ and radius r (closed at each end by plane caps normal to the axis) passing through the point P as shown in the figure. Consider a very small area ds at the point P on the Gaussian surface. iii. By Gauss' theorem, the net flux through a closed surface, $$\phi=\dfrac{q}{\varepsilon_0}\ \text{(for air/vacuum }k=1\text{)}\qquad....(1)$$ where q is the total charge inside the closed surface. iv. By symmetry, the magnitude of the electric field will be the same at all the points on the curved surface of the cylinder and will be directed radially outward. The angle between the direction of $\vec{E}$ and the normal to the curved surface of the cylinder $(\vec{ds})$ is zero. Similarly, the angle is (π/2) for the flat surface of the cylinder. i.e., $\cos\theta=1$ (for curved surface) or $\cos\theta=0$ (for flat surface) $\therefore \vec{E}\cdot\vec{ds}=Eds\cos\theta=0$ ....(For flat surface) $\therefore \vec{E}\cdot\vec{ds}=Eds\cos\theta=Eds$ ....(For curved surface) Flux dφ through the area ds, $d\phi=Eds$ Total electric flux through the Gaussian surface $$\phi=\oint \vec{E}\cdot\vec{ds}=\oint Eds=E\oint ds$$ $$\therefore \phi=E\times 2\pi rl\qquad....(2)$$ v. From equations (1) and (2) $$\dfrac{q}{\varepsilon_0}=E\times 2\pi rl$$ Since $\lambda=\dfrac{q}{l}$, $q=\lambda l$ $$\therefore \dfrac{\lambda l}{\varepsilon_0}=E2\pi rl$$ $$E=\dfrac{\lambda}{2\pi\varepsilon_0 r}$$ The direction of the electric field E is directed outward if λ is positive and inward if λ is negative.

6. 6.↑ back to the question

   Given: Q = 0.2 μC = $0.2\times10^{-6}$ C, A = 40 cm² = $40\times10^{-4}$ m², $\varepsilon_0=8.85\times10^{-12}$ S.I.units To find: i. Electric field intensity (E) ii. Mechanical force per unit area (f) Formulae: i. $E=\dfrac{\sigma}{\varepsilon_0k}$ ii. $f=\dfrac{1}{2}\varepsilon_0kE^2$ iii. $\sigma=\dfrac{Q}{A}$ Calculation: i. Using formula (i) and (iii), $$E=\dfrac{Q}{A\varepsilon_0k}=\dfrac{0.2\times10^{-6}}{40\times10^{-4}\times8.85\times10^{-12}}\quad....(k=1\text{ for air})$$ $$=\dfrac{20\times10^{-8}}{3.540\times10^{-14}}=\dfrac{20}{3.54}\times10^6$$ $$=\text{antilog}\{\log(20)-\log(3.54)\}\times10^6$$ $$=\text{antilog}\{1.3010-0.5490\}\times10^6=\text{antilog}\{0.7520\}\times10^6$$ $$=5.649\times10^6\ \text{V/m}$$ ii. From formula (ii) $$f=\dfrac{1}{2}\times8.85\times10^{-12}\times(5.649\times10^6)^2$$ $$=4.425\times10^{-12}\times(5.649)^2\times10^{12}=4.425\times(5.649)^2$$ $$=\text{antilog}\{\log(4.425)+2\log(5.649)\}=\text{antilog}\{0.6459+2(0.7520)\}=\text{antilog}\{2.1499\}$$ $$\therefore f=141.2\ \text{N/m}^2$$ Ans: i. Electric intensity is $5.649\times10^6$ V/m ii. Mechanical force per unit area is 141.2 N/m²

7. 7.↑ back to the question

   Given: R = 15 cm = 0.15 m, q = 2 μC = $2\times10^{-6}$ C, r = 20 cm = 0.2 m To find: Electric field intensity (E) Formula: $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ Calculation: From formula, $$E=\dfrac{9\times10^9\times2\times10^{-6}}{(0.2)^2}=4.5\times10^5\ \text{N/C}$$ Ans: The intensity of electric field at distance 20 cm from the centre of the ball is $4.5\times10^5$ N/C.