Which among the curves shown in Figureb cannot possibly represent electrostatic field lines?
$(a)$ The field lines showed in $(a)$ do not represent electrostatic field lines because field lines must be normal to the surface of the conductor.
$(b)$ The field lines showed in $(b)$ do not represent electrostatic field lines because the field lines cannot emerge from a negative charge and cannot terminate at a positive charge.
$(c)$ The field lines showed in $(c)$ represent electrostatic field lines. This is because the field lines emerge from the positive charges and repel each other.
$(d)$ The field lines showed in $(d)$ do not represent electrostatic field lines because the field lines should not intersect each other.
$(e)$ The field lines showed in $(e)$ do not represent electrostatic field lines because closed loops are not formed in the area between the field lines.
The given figure gives electric lines of force due to two charges $q_1$ and $q_2$. What are the signs of the two charges?
A circular disc of radius $R$ carries surface charge density $\sigma(r)=\sigma_0\left(1-\frac{r}{R}\right)$, where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is. . . . . .
A cubical volume is bounded by the surfaces $x =0, x = a , y =0, y = a , z =0, z = a$. The electric field in the region is given by $\overrightarrow{ E }= E _0 \times \hat{ i }$. Where $E _0=4 \times 10^4 NC ^{-1} m ^{-1}$. If $a =2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$. The value of $Q$ is $...........$
Take $\left.\varepsilon_0=9 \times 10^{-12} C ^2 / Nm ^2\right)$
A point charge $ + q$ is placed at the centre of a cube of side $L$. The electric flux emerging from the cube is
Explain the electric field lines and the magnitude of electric field.