A uniformly charged disc of radius $R$ having surface charge density $\sigma$ is placed in the ${xy}$ plane with its center at the origin. Find the electric field intensity along the $z$-axis at a distance $Z$ from origin :-

The electric field due to a charge at a distance of $3\, m$ from it is $500\, N/coulomb$. The magnitude of the charge is…….$\mu C$ $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N – {m^2}}}{{coulom{b^2}}}} \right]$

The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass and charge of an electron respectively are $9.1 \times 10^{-31}\,kg$ and $1.6 \times$ $10^{-19}\,C$.)

Five point charge each having magnitude $‘q’$ are placed at the corner of hexagon as shown in fig. Net electric field at the centre $‘O’$ is $\vec E$. To get net electric field at $‘O’$ be $6\vec E$, charge placed on the remaining sixth corner should be

Two point charges $( + Q)$ and $( – 2Q)$ are fixed on the $X-$axis at positions $a$ and $2a$ from origin respectively. At what positions on the axis, the resultant electric field is zero