$(a)$ Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
$\left( E _{2}- E _{1}\right) \cdot \hat{ n }=\frac{\sigma}{\varepsilon_{0}}$
where $\hat{ n }$ is a unit vector normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\hat { n }$ is from side $1$ to side $2 .$ ) Hence, show that just outside a conductor, the electric field is $\sigma \hat{ n } / \varepsilon_{0}$
$(b)$ Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
$(a)$ Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
$\left( E _{2}- E _{1}\right) \cdot \hat{ n }=\frac{\sigma}{\varepsilon_{0}}$
where $\hat{ n }$ is a unit vector normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\hat { n }$ is from side $1$ to side $2 .$ ) Hence, show that just outside a conductor, the electric field is $\sigma \hat{ n } / \varepsilon_{0}$
$(b)$ Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
$(a)$ Electric field on one side of a charged body is $E_{1}$ and electric field on the other side of the same body. is $E_z$. If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,
$\overline{E_{1}}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$
Where,
$\hat{n}=$ Unit vector normal to the surface at a point
$\sigma=$ Surface charge density at that point Electric field due to the other surface of the charged body,
$\overrightarrow{E_{2}}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$
Electric field at any point due to the two surfaces,
$\overrightarrow{E_{2}}-\overrightarrow{E_{1}}=\frac{\sigma}{2 \epsilon_{0}} \hat{n}+\frac{\sigma}{2 \epsilon_{0}} \hat{n}=\frac{\sigma}{\epsilon_{0}} \hat{n}$
$(\overrightarrow{E_{2}}-\overrightarrow{E_{1}}) \cdot \hat{n}=\frac{\sigma}{\epsilon_{0}}$
since inside a closed conductor, $\overline{E_{1}}=0$
$\therefore d \vec{E}=\overrightarrow{E_{2}}=-\frac{\sigma}{2 \epsilon_{0}} \hat{n}$
Therefore, the electric field just outside the conductor is $\frac{\sigma}{\epsilon_{0}} \hat n$
$(b)$ When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.
Similar Questions
An infinite plane sheet of charge having uniform surface charge density $+\sigma_5 \mathrm{C} / \mathrm{m}^2$ is placed on $\mathrm{x}-\mathrm{y}$ plane. Another infinitely long line charge having uniform linear charge density $+\lambda_e \mathrm{C} / \mathrm{m}$ is placed at $z=4 \mathrm{~m}$ plane and parallel to $y$-axis. If the magnitude values $\left|\sigma_s\right|=2\left|\lambda_{\mathrm{e}}\right|$ then at point $(0,0,2)$, the ratio of magnitudes of electric field values due to sheet charge to that of line charge is $\pi \sqrt{\mathrm{n}}: 1$. The value of $n$ is
An infinite plane sheet of charge having uniform surface charge density $+\sigma_5 \mathrm{C} / \mathrm{m}^2$ is placed on $\mathrm{x}-\mathrm{y}$ plane. Another infinitely long line charge having uniform linear charge density $+\lambda_e \mathrm{C} / \mathrm{m}$ is placed at $z=4 \mathrm{~m}$ plane and parallel to $y$-axis. If the magnitude values $\left|\sigma_s\right|=2\left|\lambda_{\mathrm{e}}\right|$ then at point $(0,0,2)$, the ratio of magnitudes of electric field values due to sheet charge to that of line charge is $\pi \sqrt{\mathrm{n}}: 1$. The value of $n$ is
- [JEE MAIN 2024]
An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho $. It has a spherical cavity of radius $R/2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2R$ from the axis of the cylinder, is given by the expression $\frac{{23\rho R}}{{16K{\varepsilon _0}}}$ .The value of $K$ is
An infinitely long solid cylinder of radius $R$ has a uniform volume charge density $\rho $. It has a spherical cavity of radius $R/2$ with its centre on the axis of the cylinder, as shown in the figure. The magnitude of the electric field at the point $P$, which is at a distance $2R$ from the axis of the cylinder, is given by the expression $\frac{{23\rho R}}{{16K{\varepsilon _0}}}$ .The value of $K$ is
The electric field at a distance $r$ from the centre in the space between two concentric metallic spherical shells of radii $r_1$ and $r_2$ carrying charge $Q_1$ and $Q_2$ is $(r_1 < r < r_2)$
The electric field at a distance $r$ from the centre in the space between two concentric metallic spherical shells of radii $r_1$ and $r_2$ carrying charge $Q_1$ and $Q_2$ is $(r_1 < r < r_2)$
- [AIIMS 2009]
Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius $R$, with distance $r$ from the centre $O$ is represented by:
Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius $R$, with distance $r$ from the centre $O$ is represented by:
- [JEE MAIN 2023]
The electric field $\vec E = {E_0}y\hat j$ acts in the space in which a cylinder of radius $r$ and length $l$ is placed with its axis parallel to $y-$ axis. The charge inside the volume of cylinder is
The electric field $\vec E = {E_0}y\hat j$ acts in the space in which a cylinder of radius $r$ and length $l$ is placed with its axis parallel to $y-$ axis. The charge inside the volume of cylinder is