For the given arrangement,where four charges are fixed at the corners of a square as shown,find the value of the additional charge $Q$ to be placed on one of the vertices so that the component of the net electric field along the vertical symmetric axis is zero at every point on the vertical axis.

Electric field strength due to a point charge of $5\,\mu C$ at a distance of $80\, cm$ from the charge is:

Two point charges $-10 \mu C$ and $+5 \mu C$ are situated on the $X$-axis at $x=0$ and $x=\sqrt{2} \ m$. The point along the $X$-axis where the electric field becomes zero is:

$A$ tiny $0.50\, g$ ball carries a charge of magnitude $10\, \mu C$. It is suspended by a thread in a downward electric field of intensity $300\, N/C$. If the charge on the ball is positive,then the tension in the string 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 $.......\,\mu C$ $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N \cdot m^2}}{{C^2}}} \right]$

$A$ point charge of $50 \mu C$ is placed in the $XY$ plane at a location with radius vector $\vec{r}_0 = 2 \hat{i} + 3 \hat{j} \ m$. The electric field strength magnitude at a point with radius vector $\vec{r} = 8 \hat{i} - 5 \hat{j} \ m$ is (Given: $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \ m^2 \ C^{-2}$). (in $kV \ m^{-1}$)