Consider a uniform spherical charge distribution of radius $R_1$ centred at the origin $O$. In this distribution, a spherical cavity of radius $R_2$, centred at $P$ with distance $O P=a=R_1-R_2$ (see figure) is made. If the electric field inside the cavity at position $\overrightarrow{ r }$ is $\overrightarrow{ E }(\overrightarrow{ r })$, then the correct statement$(s)$ is(are) $Image$
Consider a uniform spherical charge distribution of radius $R_1$ centred at the origin $O$. In this distribution, a spherical cavity of radius $R_2$, centred at $P$ with distance $O P=a=R_1-R_2$ (see figure) is made. If the electric field inside the cavity at position $\overrightarrow{ r }$ is $\overrightarrow{ E }(\overrightarrow{ r })$, then the correct statement$(s)$ is(are) $Image$
- [IIT 2015]
- A
$\vec{E}$ is uniform, its magnitude is independent of $R_2$ but its direction depends on $\vec{r}$
- B
$\vec{E}$ is uniform, its magnitude depends on $R_2$ and its direction depends on $\overrightarrow{ r }$
- C
$\overrightarrow{ E }$ is uniform, its magnitude is independent of $a$ but its direction depends on a $\vec{a}$
- D
$\overrightarrow{ E }$ is uniform and both its magnitude and direction depend on $\overrightarrow{ a }$
Similar Questions
$(a)$ 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.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $E$ without using Gauss’s law.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density $E$ without using Gauss’s law.
The electric field at a distance $\frac{3R}{2}$ from the centre of a charged conducting spherical shell of radius $R$ is $E.$ The electric field at a distance $\frac{R}{2}$ from the centre of the sphere is
The electric field at a distance $\frac{3R}{2}$ from the centre of a charged conducting spherical shell of radius $R$ is $E.$ The electric field at a distance $\frac{R}{2}$ from the centre of the sphere is
- [AIPMT 2010]
Mention applications of Gauss’s law.
Mention applications of Gauss’s law.
An early model for an atom considered it to have a positively charged point nucleus of charge $Ze$, surrounded by a uniform density of negative charge up to a radius $R$. The atom as a whole is neutral. For this model, what is the electric field at a distance $r$ from the nucleus?
An early model for an atom considered it to have a positively charged point nucleus of charge $Ze$, surrounded by a uniform density of negative charge up to a radius $R$. The atom as a whole is neutral. For this model, what is the electric field at a distance $r$ from the nucleus?