Physics Derivations PDF

Here is the conversion of the text in the image to markdown format: ### 1. Express Coulomb's Law in Vector Form! - $F_{12} \frac{\vec{r_1}-\vec{r_2}}{|r_1-r_2|}$ $F_{12} = \frac{kq_1q_2(r_1-r_2)}{|r_1-r_2|^2}$ = $\frac{kq_1q_2}{|r_1-r_2|^2}$ $\frac{\vec{r_1}-\vec{r_2}}{|r_1-r_2|}$ $\frac{|r_1-r_2|}...

Here is the conversion of the text in the image to markdown format: ### 1. Express Coulomb's Law in Vector Form! - $F_{12} \frac{\vec{r_1}-\vec{r_2}}{|r_1-r_2|}$ $F_{12} = \frac{kq_1q_2(r_1-r_2)}{|r_1-r_2|^2}$ = $\frac{kq_1q_2}{|r_1-r_2|^2}$ $\frac{\vec{r_1}-\vec{r_2}}{|r_1-r_2|}$ $\frac{|r_1-r_2|}{|r_1-r_2|^2}$. Force applied by $F_{21}$: - = $F_{12} = \frac{kq_1q_2(r_1-r_2)}{|r_1-r_2|^3}$ $\vec{F_{21}} = \frac{kq_1q_2(\vec{r_2}-\vec{r_1})}{|r_2-r_1|^2}$ = $\frac{kq_1q_2}{|r_2-r_1|^2}$ $(\vec{r_2}-\vec{r_1})$ $\vec{F_{21}} = \frac{kq_1q_2(\vec{r_2}-\vec{r_1})}{|r_2-r_1|^3}$ $F_{21} = -F_{12}$ ### 2. Derive an expression for Electric Field at any point on axial line of the Electric Dipole:- The image appears to show a diagram illustrating an electric dipole along the x-axis with charges -q and +q separated by a distance 2l. There's a point P at a distance x from the center of the dipole where the electric field needs to be calculated. Distances are $x+l$ and $x-l. Net Electric field at point P! $\vec{E_{net}} = \vec{E_2} - \vec{E_1}$ Due to +q, $E_2 = \frac{kq}{(x-l)^2}$ Due to -q, $E_1 = \frac{kq}{(x+l)^2}$ $E_{net} = \frac{kq}{(x-l)^2} - \frac{kq}{(x+l)^2} = kq[\frac{1}{(x-l)^2} - \frac{1}{(x+l)^2}]$ $= kq[\frac{(x+l)^2 - (x-l)^2}{(x^2 - l^2)^2}] = kq [\frac{4xl}{(x^2 - l^2)^2}] = \frac{kq(2l)(2x)}{(x^2 - l^2)^2} [ \because p = q \times 2l]$ $\vec{E} = \frac{2kpx}{(x^2 - l^2)^2}$ For ideal dipole! - $l << r$, so $l^2 \rightarrow neg.$ $E = \frac{2kp}{x^3}$ or $E = \frac{1}{4\pi\epsilon_0} \frac{2p}{x^3}$ ### 3. Derive an expression for Electric field at a point on Equitorial line of an Electric dipole:- The image shows the geometry for calculating the electric field at a point P on the equatorial line of an electric dipole. The dipole consists of charges -q and +q separated by a distance 2l. Point P is at a distance r from the center of the dipole. $r^2 = x^2 + l^2$ $r = \sqrt{x^2+l^2}$ Here, $E_1=E_2=E$ Due to -q &+q E$_1 = \frac{kq}{x^2 + l^2}$ $E_2 = \frac{kq}{x^2 + l^2}$ Thus, Net Field: - $E_{net} = 2Ecos\theta = 2E[\frac{l}{r}]$ $[cos\theta = \frac{1}{r}]$ $= 2\frac{kq}{x^2+l^2} \times \frac{l}{\sqrt{x^2+l^2}} = \frac{kp}{(x^2+l^2)^{3/2}}$ For ideal dipole, $l << r$, $l^2 \rightarrow 0$ $E = \frac{kp}{x^3}$ or $E = \frac{1}{4\pi\epsilon_0} \frac{p}{x^3}$ ### 4. Deduce the Expression for torque acting on a dipole of dipole moment $\vec{p}$ in presence of Uniform Electric field ($\vec{E}$): - :: Torque $= f \times d$ where, d = Perpendicular dist. $\vec{\tau} = \vec{\tau_1} + \vec{\tau_2} = f \times d + f \times d = qE(lsin\theta) + qE(lsin\theta) = 2qElsin\theta$ $[\because P= q \times 2l]$ or $\tau = pEsing$ $\tau = \vec{p} \times \vec{E}$ $\theta = \text{Angle bet} \vec{p} \times \vec{E} | \vec{p} \& \vec{E}$ Case I When $\theta = 0$ $\tau = pEsin\theta = pEsin0^\circ$ $\tau = 0$ Stable Equilibrium Case II When $\theta = 180^\circ$ $\tau = pEsin\theta = pEsin180^\circ$ $\tau = 0E$ Unstable Equilibrium Case Ⅲ When $\theta = 90^\circ$ $\tau = pEsin\theta = pEsin90^\circ$ $\tau = pE$ max. Torque ### 5. Derive Gauss Law:- The net Electric field through a closed surface (3D) is $\frac{1}{\epsilon_0}$ times then net charge enclosed by surface. $\phi_{closed} = \frac{q_{in}}{\epsilon_0} = \oint \vec{E}.d\vec{A}$ Verification of Gauss law using Coulomb's law: - $\therefore$ Electric flux:- $d\phi = E.dA \text{d} \phi = E d A cos \theta $ [$\theta = 0^\circ$] $d\phi = E dA$ Since $\vec{E}$ is same everywhere, on spherical surface:- $\int d\phi = \int E. dA$ $\phi = EA$ where, $E = \frac{1}{4\pi\epsilon_0 R^2}$ & $A = 4\pi R^2$ $\phi = \frac{1}{4\pi\epsilon_0}\frac{q}{R^2} 4\pi R^2$ $\phi = \frac{q}{\epsilon_0}$ ### 6. Derive expression for Electric field ($\vec{E}$) due to a straight uniformly charged infinite line of charge density. Here $\lambda = \text{Linear charge density}$, $\lambda = \frac{q}{r}$ Here, Gaussian surface is cylinderical & The direction of $\vec{E}$ field is radially outward because of $+$ive charge. The image shows a diagram of an infinite line charge with Gaussian surfaces around it to calculate the electric field using Gauss's law. I'm not certain what to replace the variable 'Dine' with, so I left a question mark. $\phi_1 = \int E dA cos\theta [\theta = 90^\circ$] $\phi_2 = \int E dA cos\theta [\theta = 90^\circ$] $\phi_3 = \int E dA cos\theta [\theta = 0^\circ$] Net flux! $\phi_{het} = \phi_1 + \phi_2 + \phi_3$ $= \int EdAcos90 + \int EdAcos90 + \int EdAcos0^\circ = 0 + 0 + \int E dA$ $\phi = EA$ Total Curved area of surface $= 2 \pi r l$ so $ \phi = E(2\pi r l)$ Acc. to Gauss law. $\phi = \frac{q_{in}}{\epsilon_0}$ $E (2\pi r l) = \frac{\lambda l}{\epsilon_0} $ [$q_{in} = \lambda l$] $E = \frac{\lambda}{2\pi r \epsilon_0}$ ### 7. Find Electric field due to uniformly charged infinity large plane thin sheet with surface charge density. $\sigma$ = Surface charge density; Gaussian surface $\rightarrow$ Cylinderical. (The image is of a pillbox) Tolal flux! - The image appears to show a cylindrical Gaussian surface intersecting a charged plane. $\phi_{net} = \phi_1 + \phi_2 + \phi_3 = \int E dAcos0^\circ + \int E dAcos0^\circ + \int E dAcos90^\circ $ $= \int E dA + \int E dA + 0$ += EA + EA $\phi = 2EA \qquad (1)$ According to Gauss law' - $\phi = \frac{qin}{\epsilon_0} \qquad (2)$ $2EA= \frac{\sigma A}{\epsilon_0} \qquad [q = \sigma A]$ $E=\frac{\sigma}{2 \epsilon_0}$ It is independent of $r$. ### 8. Derive the Expression for potential due to point charge! The image is of a positive charge at a distance from another infinitesimally small charge. Here, P point at a distance, $r$ from +q charge Workdone to bing $q_0$ from infinite to point P is equal to! - $W(p \rightarrow \infty) = \int_p^{\infty} F_{ex} dr = \int_p^{\infty} \frac{Kq_0 q}{r^2} dr cos\theta$ $= -Kq q_0 \int_{\infty}^r \frac{1}{r^2} dr [cos \theta = -1]$ $W(p\rightarrow \infty) = \frac{K q q_0}{r}$ Electric Potential:- $V = \frac{W_{ex}}{q_0} = \frac{Kq q_0}{r} \times \frac{1}{q} = \frac{K q}{r}$ or $V = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r}$ ### (9) Derive the expression for the Electric potential due to an Electric dipole at any point on the axial line. We know that electric potential will be!: $V=\frac{KQ}{r}$ Net Potential!: $V_{net} = V_1+V_2$ $V_{net}=\frac{-Kq}{(x+l)}+\frac{Kq}{(x-l)}$ $=Kq[\frac{1}{x-l}-\frac{1}{x+l}]$ $=Kq[\frac{x+l-(x-l)}{x^2-l^2}]$ $=Kq(\frac{2l}{x^2-l^2})$ $Vnet = \frac{KP}{x^2 - l^2}$ Generally x>>>l, $l^2 \rightarrow 0$ $Vnet = \frac{KP}{x^2}$ or $V_{net}= \frac{1}{4π\epsilon_0} \frac{P}{x^2}$ * Electric Potential at any Point ### (10) Potential Energy of an Electric dipole in Uniform Electric field. Asw e know that In uniform electric field dipole experiences net force! :$\tau = \vec{P}\times\vec{E}$ $(\tau=PE\sin \theta)$ Net Workdone in rotating dipole. - $W_{el}=\int_{\theta_1}^{\theta_2} \tau . d\theta$ $=\int_{\theta_1}^{\theta_2} PE\sin\theta .d\theta$ $=-\int_{\theta_1}^{\theta_2} PE d(\cos \theta)$ $=-PE[\cos \theta]_{\theta_1}^{\theta_2}$ $W_{el} = +PE[\cos \theta ^2-\cos\theta 1]$ By potential energy: - $du=-Wex$ $U_2-U_1= PE (\cos \theta^2-\cos\theta1)$ Put, $\theta_=90$ & $U_1=0$ $U = -PECos$ $U=-\vec{P}\vec{E}$ Case(i) $\theta=90^\circ$ $\tau=PEsin90$ $= PE$. $U=-PECos0^\circ =- PECos90$ $=PE$. $W_{el}=+ PE[\cos 90-\cos 0]=- PE$ $V_2=PE$ Case (ii) $\sigma =\sigma,0, 0+180^\circ$ $\tau= V_2=0 PEsino$ $\qquad \text{and} \qquad \tau=PEsin\theta$ $U=-PECos o \qquad \text{and} \qquad =-PEsonlso$ $U=-PE$ $V=PE\left( \frac{q}{v} = c\right)$ ### 10 Deduce the expression for Energy stored in Capacitor & also find the Energy density: The amount of workdone to add charge to a capacitor is stored in form of electric potential energy in the spare between the 2 blades. Let 'g' is the charge, given to a capacitor &v be the potential difference. Add dq charge to a capacitor. Workdone $dw = dqv$ Integrate it!: $\int_{0}^{w} dw = \int_{0}^{q} dq v$ $w=\int_{0}^{q} \frac{q}{L} dq$ $w= [(9^2/ 2]/c]^q_0= [ q^2-0](1/c)$ Potential Energy: $U=q^2/(2c)$ $U= 1/2 (v)^2/ c=(1c^2v^2)/(2c)\qquad =LV^4=0 $ ### U= 1/2 (Q/V.2) $UU = \frac{1}{2} Qv \text{(energy/ volume)}$ Energy Density = U= 1CV^{2}/2 / Ad = 1 epsilon A * E^2 d /2 /Ad = epslm^2 0 u = epsilon^2 . ( epsilon 0 / 2] $$u =\frac12 \epsilon {0E^2} $$ ### 11 Obtain the Expression of Drift velocity? At any instant of time, average thermal velocity of 𝑒−in a, electron is zero: - $V_{av}=0, V_1+V_2+V_3+.....+_n/n$