(D) hollow conducting sphere is an equipotential volume. Since the entire material of the conductor is at the same potential,the potential at any poin

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(D) hollow conducting sphere is an equipotential volume. Since the entire material of the conductor is at the same potential,the potential at any point on the surface ($A$ and $B$) and any point inside the cavity $(C)$ must be equal. Therefore,$V_A = V_B = V_C$.

Class 12 Physics Electric Potential and Capacitance Conductor, Electrostatic Shielding, Induced Charge and Charge Redistribution on conductor

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$A$ hollow conducting sphere is placed in an electric field produced by a point charge placed at $P$ as shown in the figure. Let $V_A, V_B$,and $V_C$ be the potentials at points $A, B$,and $C$ respectively,where $A$ and $B$ are on the surface of the sphere and $C$ is inside the sphere. Then:

A
$V_C > V_B$
B
$V_B > V_C$
C
$V_A > V_B$
D
$V_A = V_B = V_C$

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Electric Potential and Capacitance

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Conductor, Electrostatic Shielding, Induced Charge and Charge Redistribution on conductor

$A$ and $B$ are two identical spheres. The charges on them are $7\,\mu C$ and $1\,\mu C$ respectively. Now,both are connected by a wire. Calculate the flow of charge from $A$ to $B$ in $\mu C$.

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$A$ charge $+q$ is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius $R_1$ and outer radius $R_2$. $A$ charge $+Q$ is placed at a distance $r > R_2$ from the centre of the shell. Then the electric field in the hollow cavity

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$A$ solid conducting sphere of radius $a$ has a net positive charge $2Q$. $A$ conducting spherical shell of inner radius $b$ and outer radius $c$ is concentric with the solid sphere and has a net charge $-Q$. The surface charge density on the inner and outer surfaces of the spherical shell will be

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'At the surface of a charged conductor,the electrostatic field must be normal to the surface at every point'. Explain.

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Two conducting spheres $A$ and $B$ have radii $1 \, mm$ and $2 \, mm$ respectively and are charged. They are placed at a distance of $5 \, cm$. When they are connected by a conducting wire,the ratio of the electric field intensities on their surfaces at equilibrium is:

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$A$ hollow conducting sphere is placed in an electric field produced by a point charge placed at $P$ as shown in the figure. Let $V_A, V_B$,and $V_C$ be the potentials at points $A, B$,and $C$ respectively,where $A$ and $B$ are on the surface of the sphere and $C$ is inside the sphere. Then:

Similar Questions

$A$ and $B$ are two identical spheres. The charges on them are $7\,\mu C$ and $1\,\mu C$ respectively. Now,both are connected by a wire. Calculate the flow of charge from $A$ to $B$ in $\mu C$.

$A$ charge $+q$ is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius $R_1$ and outer radius $R_2$. $A$ charge $+Q$ is placed at a distance $r > R_2$ from the centre of the shell. Then the electric field in the hollow cavity

$A$ solid conducting sphere of radius $a$ has a net positive charge $2Q$. $A$ conducting spherical shell of inner radius $b$ and outer radius $c$ is concentric with the solid sphere and has a net charge $-Q$. The surface charge density on the inner and outer surfaces of the spherical shell will be

'At the surface of a charged conductor,the electrostatic field must be normal to the surface at every point'. Explain.

Two conducting spheres $A$ and $B$ have radii $1 \, mm$ and $2 \, mm$ respectively and are charged. They are placed at a distance of $5 \, cm$. When they are connected by a conducting wire,the ratio of the electric field intensities on their surfaces at equilibrium is:

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