When electric flux is said to be positive, negative or zero ?
When electric flux is said to be positive, negative or zero ?
Similar Questions
Which of the following figure represents the electric field lines due to a single positive charge?
Which of the following figure represents the electric field lines due to a single positive charge?
Using thomson's model of the atom, consider an atom consisting of two electrons, each of charge $-e$, embeded in a sphere of charge $+2e$ and radius $R$. In equilibrium each electron is at a distance $d$ from the centre of the atom. What is the equilibrium separation between electrons
Using thomson's model of the atom, consider an atom consisting of two electrons, each of charge $-e$, embeded in a sphere of charge $+2e$ and radius $R$. In equilibrium each electron is at a distance $d$ from the centre of the atom. What is the equilibrium separation between electrons
A charge $Q$ is placed at a distance $a/2$ above the centre of the square surface of edge $a$ as shown in the figure. The electric flux through the square surface is
A charge $Q$ is placed at a distance $a/2$ above the centre of the square surface of edge $a$ as shown in the figure. The electric flux through the square surface is
- [JEE MAIN 2018]
A charged shell of radius $R$ carries a total charge $Q$. Given $\Phi$ as the flux of electric field through a closed cylindrical surface of height $h$, radius $r$ and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct ? $\epsilon_0$ is the permittivity of free space]
$(1)$ If $h >2 R$ and $r > R$ then $\Phi=\frac{ Q }{\epsilon_0}$
$(2)$ If $h <\frac{8 R }{5}$ and $r =\frac{3 R }{5}$ then $\Phi=0$
$(3)$ If $h >2 R$ and $r =\frac{4 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$
$(4)$ If $h >2 R$ and $r =\frac{3 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$
A charged shell of radius $R$ carries a total charge $Q$. Given $\Phi$ as the flux of electric field through a closed cylindrical surface of height $h$, radius $r$ and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct ? $\epsilon_0$ is the permittivity of free space]
$(1)$ If $h >2 R$ and $r > R$ then $\Phi=\frac{ Q }{\epsilon_0}$
$(2)$ If $h <\frac{8 R }{5}$ and $r =\frac{3 R }{5}$ then $\Phi=0$
$(3)$ If $h >2 R$ and $r =\frac{4 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$
$(4)$ If $h >2 R$ and $r =\frac{3 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$
- [IIT 2019]
The inward and outward electric flux for a closed surface in units of $N{\rm{ - }}{m^2}/C$ are respectively $8 \times {10^3}$ and $4 \times {10^3}.$ Then the total charge inside the surface is [where ${\varepsilon _0} = $ permittivity constant]
The inward and outward electric flux for a closed surface in units of $N{\rm{ - }}{m^2}/C$ are respectively $8 \times {10^3}$ and $4 \times {10^3}.$ Then the total charge inside the surface is [where ${\varepsilon _0} = $ permittivity constant]