Five point charges each having magnitude $q$ are placed at the corners of a regular hexagon as shown in the figure. The net electric field at the centre $O$ is $\vec E$. To make the net electric field at $O$ equal to $6\vec E$,the charge placed on the remaining sixth corner should be: (in $,q$)

$A$ thin metallic wire having a cross-sectional area of $10^{-4} \, m^2$ is used to make a ring of radius $30 \, cm$. $A$ positive charge of $2 \pi \, pC$ is uniformly distributed over the ring, while another positive charge of $30 \, pC$ is kept at the centre of the ring. The tension in the ring is . . . . . . $N$; provided that the ring does not get deformed (neglect the influence of gravity). (Given, $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, SI$ units)

Two point charges $-q$ and $+q/2$ are situated at the origin $(0, 0, 0)$ and at the point $(a, 0, 0)$ respectively. The point along the $X$-axis where the electric field vanishes is:

Two charged particles, each with a charge of $+q$, are located along the $x$-axis at $x = 2$ and $x = 4$, as shown below. Which of the following shows the graph of the magnitude of the electric field along the $x$-axis from the origin to $x = 6$?

Two negative charges of equal magnitude are located in the $xy$-plane as shown in the figure. The direction of the electric field at point $P$ is

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