Within a spherical charge distribution of charge density $\rho \left( r \right)$, $N$ equipotential surfaces of potential ${V_0},{V_0} + \Delta V,{V_0} + 2\Delta V,$$.....{V_0} + N\Delta V\left( {\Delta V > 0} \right),$ are drawn and have increasing radii $r_0, r_1, r_2,......r_N$, respectively. If the difference in the radii of the surfaces is constant for all values of $V_0$ and $\Delta V$ then
Within a spherical charge distribution of charge density $\rho \left( r \right)$, $N$ equipotential surfaces of potential ${V_0},{V_0} + \Delta V,{V_0} + 2\Delta V,$$.....{V_0} + N\Delta V\left( {\Delta V > 0} \right),$ are drawn and have increasing radii $r_0, r_1, r_2,......r_N$, respectively. If the difference in the radii of the surfaces is constant for all values of $V_0$ and $\Delta V$ then
- [JEE MAIN 2016]
- A
$\rho \left( r \right) = $ constant
- B
$\rho \left( r \right) \propto \frac{1}{{{r^2}}}$
- C
$\rho \left( r \right) \propto \frac{1}{r}$
- D
$\rho \left( r \right) \propto r$
Similar Questions
The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5{x^2} + 10x - 9)\,volt$. Value of electric field at $x = 1$ is......$V/m$
The electric potential $V$ is given as a function of distance $x$ (metre) by $V = (5{x^2} + 10x - 9)\,volt$. Value of electric field at $x = 1$ is......$V/m$
The electric potential at a point $(x,\;y)$ in the $x - y$ plane is given by $V = - kxy$. The field intensity at a distance $r$ from the origin varies as
The electric potential at a point $(x,\;y)$ in the $x - y$ plane is given by $V = - kxy$. The field intensity at a distance $r$ from the origin varies as
A spherical charged conductor has surface charge density $\sigma $ . The electric field on its surface is $E$ and electric potential of conductor is $V$ . Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be
A spherical charged conductor has surface charge density $\sigma $ . The electric field on its surface is $E$ and electric potential of conductor is $V$ . Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be
Two metal pieces having a potential difference of $800 \;V$ are $0.02\; m$ apart horizontally. A particle of mass $1.96 \times 10^{-15} \;kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge, then charge on the particle is
Two metal pieces having a potential difference of $800 \;V$ are $0.02\; m$ apart horizontally. A particle of mass $1.96 \times 10^{-15} \;kg$ is suspended in equilibrium between the plates. If $e$ is the elementary charge, then charge on the particle is
$A B C$ is a right angled triangle situated in a uniform electric field $\vec{E}$ which is in the plane of the triangle. The points $A$ and $B$ are at the same potential of $15 \,V$ while the point $C$ is at a potential of $20 \,V . A B=3 \,cm$ and $B C=4 \,cm$. The magnitude of electric field is (in $S.I.$ Units)
$A B C$ is a right angled triangle situated in a uniform electric field $\vec{E}$ which is in the plane of the triangle. The points $A$ and $B$ are at the same potential of $15 \,V$ while the point $C$ is at a potential of $20 \,V . A B=3 \,cm$ and $B C=4 \,cm$. The magnitude of electric field is (in $S.I.$ Units)