Electrostatic field satisfy

$\oint \vec E \vec{dl} = 0$

$\oint \vec{E} \cdot \vec{dl}=0$

$\oint \vec{E} \cdot \overrightarrow{d l}=\int_A^B \vec{E} \cdot \vec{dl}+\int_B^C \vec{E} \cdot \vec{dl} +\int_C^D \vec{E} \cdot \vec{dl}+\int_D^A \vec{E} \cdot \vec{dl}$

$\int_A^B \vec{E} \cdot \overrightarrow{d l}=\int E_0 x \hat{j} \cdot \hat{\imath} d x = 0$

$\int_C^D \vec{E} \cdot \vec{dl}=0$

$\int_{B}^{C} \vec E \cdot \vec{dl} = \int_{0}^{b} E_0 a \hat{j} \cdot \hat{j} d y = \int_{0}^{b} E_0 a d y = E_0 a b$

$\int_D^A \vec{E} \cdot \vec{dl}=0$

$\oint \vec{E} \cdot \vec{dl}=E_0 a b \neq 0$

$\vec{E}=E_0 x \hat{\jmath}$ cannot represent an electrostatic field.

How about the following fields.

$\vec{E}=E_0 x \hat{\imath}$

Can they represent an electrostatic fields?

Given the following electrostatic potential, calculate the corresponding electric field.

$v =v_0 ; \quad r=\sqrt{x^2+y^2+z^2}

$E_x=-\frac{\partial v}{\partial x}$

$E_y=-\frac{\partial v}{\partial y}$

$E_z=-\frac{\partial v}{\partial z}$

A.

For $r

$E_x=-\frac{\partial v}{\partial x}=0$

$E_y=-\frac{\partial v}{\partial y}=0$

$E_z=-\frac{\partial v}{\partial z}=0$

B. For $x>a$

$E_x =-\frac{\partial v}{\partial x}$

$=-\frac{\partial}{\partial x}\left[\frac{v_0 a}{\sqrt{x^2+y^2+z^2}}\right] =-v_0 a \frac{\partial}{\partial x}\left(\frac{1}{\sqrt{x^2+y^2 z^2}}\right)$

$=-v_0 a\left(-\frac{1}{2}\right) \frac{1}{\left(x^2+y^2+z^2\right)^{3 / 2}}(2 x) =\frac{v_0 a x}{\left(x^2+y^2+z^2\right)^{7 / 2}}=\frac{v_0 a x}{r^7}$

$\epsilon_y =-\frac{\partial v}{\partial y}=\frac{V_0 a y}{r^3}$

$\epsilon_z =-\frac{\partial v}{\partial z}=\frac{v_0 a z}{r^3}$

$\vec{E} = E_x \hat{i}+E_y \hat{\jmath}+E_z \hat{k}$

$= \frac{v_0 a}{r^3}(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) = \frac{v_0 a \vec{r}}{r^3}$

$\vec{r} =x \hat{\imath}+y \hat{\jmath}+z \hat{k}$

$\int_P^Q \vec{E} .\vec{dl}=V(P)-V(Q)$

$V(P)=\frac{2 q}{4 \pi E_0 d/2}-\frac{2 q}{4 \pi E_0 d/2}=0$

$V(Q)=\frac{2 q}{4 \pi E_0 \cdot d / 2}-\frac{2 q}{4 \pi E_0 3d/ 2} =\frac{2 q}{3 \pi E_0 d}$

$\int_P^Q \vec{E} \cdot \vec{dl}=V(P)-V(Q)=-\frac{2 q}{3 \pi E_0 d}$

$V=-\int_A^D \vec{E} \cdot \vec{dl}$

= $-\int_A^B \bar{E} \cdot \hat{i} d x-\int_B^C \vec{E} \cdot \hat {j} d y-\int_C^D \vec{E} \cdot \hat{k} d z$

= $-\int_0^2 20 d x-\int_0^2 30 dy - 0$

= -40-60 = -100 V

$T \cos \theta=m g$

$T \sin \theta=F_e=\frac{q^2}{4 \pi E_0(2 l \sin \theta)^2}$

$=\frac{q^2}{16 \pi E_0 l^2 \sin ^2 \theta}$

$\tan \theta=\frac{q^2}{16 \pi E_0 l^2 sin^2 \theta}.\frac{1}{m g}$

$q^2=16 \pi E_0 l^2 \sin ^2 \theta \tan^2 \theta$

$\oint \vec{D} \cdot d \vec{A}= Q_{F_{enc}}$

$\vec{D}=\epsilon_0 \vec{E}+\vec{P}$

$\vec{D}, \vec{E} \ \vec{P}$ will be along radial direction and unit only depends on r .

A. For 0 < r < R

$\oint \vec{D} \cdot d \vec{A} =\alpha_{f_{enc}}$

$\text { D. } 4 \pi r^2 = \int_0^r \alpha r \cdot 4 \pi r^2 d r$

$=4 \pi \alpha \int_0^r r^3 d r =\pi \alpha r^4$

0 < r < R

$\vec{D} =\frac{\alpha}{4} r^2 \hat{r}$

$\vec{E}=\frac{\vec{D}}{\epsilon_0 K_i}=\frac{\alpha \hat{k}}{4 \varepsilon_0 k_1} \hat{r}$

$\vec{P}=\vec{D}-\epsilon_0 \vec{E}$

$=\epsilon_0\left(K_1-1\right) \vec{E} =\frac{\left(K_1-1\right)}{4 K_1} \alpha r^2 \hat{r}$

$ R

$\oint \vec{D} \cdot d \vec{A}=Q_{F_{enc}}$

$4 \pi r^2 D=\int_0^R \alpha r 4 \pi r^2 d r =\pi \alpha R^4$

$\vec{D}=\frac{\alpha}{4} \frac{R^4}{r^2} \hat{r}$

$\vec{E}=\frac{\vec{D}}{E_0 K_2}=\frac{\alpha R^4}{4 E_0 K_2 r^2} \hat{r}$

$\vec{P}=\vec{D}-\epsilon_0 \vec{E}$

$ R

$\oint \vec{D} \cdot d \vec{A}=Q_{F_{enc}}$

$4 \pi r^2 D=\int_0^R \alpha r 4 \pi r^2 d r = \pi \alpha R^4$

$\vec{D}=\frac{\alpha}{4} \frac{R^4}{r^2} \hat{r}$

$\vec{E}=\frac{\vec{D}}{E_0 K_2}=\frac{\alpha R^4}{4 E_0 K_2 r^2} \hat{r}$

$\vec{P}=\vec{D}-\epsilon_0 \vec{E}$

For r > 2R

$\oint \vec{D} \cdot d \vec{A}=Q_{F_{enc}}$

$4 \pi r^2 D = \pi \alpha R^4$

$\vec{D}=\frac{\alpha}{4} \frac{R^4}{r^2} \hat{r}$

$\vec{E}$= $\frac {\vec{D}}{E_0}$ = $\frac{\alpha R^4}{4 E_0 K_2 r^2} \hat{r}$

$\vec{P}=\vec{D}-\epsilon_0 \vec{E}$

A point charge Q is placed at a distance $\frac {a}{2}$ above the centre of a horizontal square surface of edge Q . The electrostatic flux the angle ….. square surface will be .

a) $\frac {Q}{E_0}$

b) $\frac {Q}{4E_0}$

c)$\frac {Q}{6E_0}$

d) Zero

Consider a set of two point charge $q_1$ abd $q_2$ and a uniformaly charges sphere of radius R with uniform volume charge density $\rho$ obtain the volume of .

a) $\oint \vec E. \vec {dA}$ over the closed surface S.

b) $\oint \vec E. \vec {dl}$ are curve C.

c) For what volume of $\rho$ with $\oint \vec E. \vec {dA}$ over S remain?

d) Draw a closed surface triangle what $\oint \vec E. \vec {dl}$ will be independence of $\rho$

a) $\Rightarrow \oint_S \vec{E} \cdot d \vec{A}=\frac{Q}{E_0}=\frac{1}{E_0}\left[q_2+\frac{4 \pi}{3} R^3 \rho\right]$

b) $\oint \vec{E} \cdot \overrightarrow{dl}=0$

The relationn $\vec D$ = $\epsilon_{0} \vec {E} + \vec {P}$ holds good.

a) only in free space

b) only inside a dielectric

c) only outside a dielectric

d) everywhere in space

a) $\frac {\rho_1}{\rho_2}$ = $- \frac {32}{25}$

b) $\frac {\rho_1}{\rho_2}$ = - 4