If $q$ is the charge per unit area on the surface of a conductor, then the electric field intensity at a point on the surface is
A$\left( {\frac{q}{{{\varepsilon _0}}}} \right)$ normal to surface
B$\left( {\frac{q}{{2{\varepsilon _0}}}} \right)$ normal to surface
C$\left( {\frac{q}{{{\varepsilon _0}}}} \right)$ tangential to surface
D$\left( {\frac{q}{{2{\varepsilon _0}}}} \right)$ tangential to surface
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A$\left( {\frac{q}{{{\varepsilon _0}}}} \right)$ normal to surface
a (a)
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The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then
In a certain charge distribution, all points having zero potential can be joined by a circle $S$. Points inside $S$ have positive potential and points outside $S$ have negative potential. A positive charge, which is free to move, is placed inside $S$
Two point charges $-Q$ and $+Q / \sqrt{3}$ are placed in the xy-plane at the origin $(0,0)$ and a point $(2,0)$, respectively, as shown in the figure. This results in an equipotential circle of radius $R$ and potential $V =0$ in the $xy$-plane with its center at $(b, 0)$. All lengths are measured in meters.
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation