Class 12PhysicsElectric Potential and CapacitanceConductor, Electrostatic Shielding, Induced Charge and Charge Redistribution on conductor
Easy$A$ neutral conducting solid sphere of radius $R$ has two spherical cavities of radius $a$ and $b$ as shown in the figure. The centre-to-centre distance between the two cavities is $c$. Charges $q_a$ and $q_b$ are placed at the centres of the cavities, respectively. The force between $q_a$ and $q_b$ is:
- A$\frac{1}{4 \pi \varepsilon_0} \frac{q_{a} q_{b}}{c^2}$
- B$\frac{1}{4 \pi \varepsilon_0} q_a q_b\left(\frac{1}{a^2}+\frac{1}{b^2}\right)$
- Czero
- Dinsufficient data
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Similar Questions
Assertion: $A$ positive charge particle is placed in front of a spherical uncharged conductor. The number of lines of force terminating on the sphere will be more than those emerging from it.
Reason: The surface charge density at a point on the sphere nearest to the point charge will be negative and maximum in magnitude compared to other points on the sphere.
Reason: The surface charge density at a point on the sphere nearest to the point charge will be negative and maximum in magnitude compared to other points on the sphere.
Consider three concentric metallic spheres $A, B$ and $C$ of radii $a, b, c$ respectively,where $a < b < c$. $A$ and $B$ are connected,whereas $C$ is grounded. The potential of the middle sphere $B$ is raised to $V$. Then the charge on the sphere $C$ is
DifficultKVPY 2012
View SolutionShow that electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface.
Easy
View Solution$A$ spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B < R_A$ are kept far apart and each is given charge $+Q$. Now they are connected by a thin metal wire. Then:
$(A)$ $E_A^{\text{inside}} = 0$
$(B)$ $Q_A > Q_B$
$(C)$ $\frac{\sigma_A}{\sigma_B} = \frac{R_B}{R_A}$
$(D)$ $E_A^{\text{on surface}} < E_B^{\text{on surface}}$
$(A)$ $E_A^{\text{inside}} = 0$
$(B)$ $Q_A > Q_B$
$(C)$ $\frac{\sigma_A}{\sigma_B} = \frac{R_B}{R_A}$
$(D)$ $E_A^{\text{on surface}} < E_B^{\text{on surface}}$
DifficultIIT 2011
View Solution$A$ metallic spherical shell has an inner radius $R_1$ and an outer radius $R_2$. $A$ charge $Q$ is placed at the center of the spherical cavity. What will be the surface charge density on $(i)$ the inner surface,and $(ii)$ the outer surface?
Medium
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