$ABC$ is an equilateral triangle. Charges $+q$ are placed at each corner. The electric intensity at $O$ will be

Four point positive charges of same magnitude $(Q)$ are placed at four corners of a rigid square frame of side $L$ as shown in the figure. The plane of the frame is perpendicular to the $Z$-axis. If a negative point charge $(-q)$ is placed at a small distance $z$ away from the center of the frame along the $Z$-axis $(z < < L)$, then:

Two point charges of magnitudes $-8 \mu C$ and $+32 \mu C$ are separated by a distance of $15 \ cm$ in air. The position of the point from the $-8 \mu C$ charge at which the resultant electric field becomes zero is (in $cm$)

The distance between two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$. At what point on the line joining the two charges will the electric field intensity be zero?

Two point charges $q_1 = \sqrt{10} \, \mu C$ and $q_2 = -25 \, \mu C$ are placed on the $x$-axis at $x = 1 \, m$ and $x = 4 \, m$ respectively. The electric field (in $V/m$) at a point $y = 3 \, m$ on the $y$-axis is,[ take $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, Nm^2C^{-2}$ ]

In the given figure,the electric field at the center $O$ due to section $AB$ of a uniformly charged ring is $\overrightarrow{E}$. What will be the electric field at $O$ due to the remaining section $ACB$?