Difficult
Shown in the figure is a shell made of a conductor. It has inner radius $a$ and outer radius $b$,and carries a total charge $Q$. At its centre is a dipole $\vec{P}$ as shown. In this case,
- ASurface charge density on the inner surface of the shell is zero everywhere.
- BElectric field outside the shell is the same as that of a point charge $Q$ at the centre of the shell.
- CSurface charge density on the inner surface is uniform and equal to $\frac{(Q/2)}{4\pi a^2}$.
- DSurface charge density on the outer surface depends on $|\vec{P}|$.
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Similar Questions
What will be the total flux through the faces of the cube as shown in the figure with side of length $a$ if a charge $q$ is placed at:
$(a)$ $C$: centre of a face of the cube.
$(b)$ $D$: midpoint of $B$ and $C$.
$(a)$ $C$: centre of a face of the cube.
$(b)$ $D$: midpoint of $B$ and $C$.
Medium
View Solution$A$ point charge $q$ is placed at the center $O$ of a cube of side length $L$. Another identical charge $q$ is placed at a distance $L$ from $O$ as shown in the figure. The electric flux through the face $ABCD$ is:
$A$ charge is kept at the central point $P$ of a cylindrical region. The two edges subtend a half-angle $\theta$ at $P$, as shown in the figure. When $\theta=30^{\circ}$, then the electric flux through the curved surface of the cylinder is $\Phi$. If $\theta=60^{\circ}$, then the electric flux through the curved surface becomes $\Phi / \sqrt{n}$, where the value of $n$ is. . . . . . .
Draw the electric field lines for a system of two identical positive charges placed near each other.
Medium
View Solution$A$ few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that:
$(A)$ $|Q_1| > |Q_2|$
$(B)$ $|Q_1| < |Q_2|$
$(C)$ at a finite distance to the left of $Q_1$ the electric field is zero
$(D)$ at a finite distance to the right of $Q_2$ the electric field is zero
$(A)$ $|Q_1| > |Q_2|$
$(B)$ $|Q_1| < |Q_2|$
$(C)$ at a finite distance to the left of $Q_1$ the electric field is zero
$(D)$ at a finite distance to the right of $Q_2$ the electric field is zero
MediumIIT 2010
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