Two point charges $q_1$ and $q_2 (=q_1/2)$ are placed at points $A(0, 1)$ and $B(1, 0)$ as shown in the figure. The electric field vector at point $P(1, 1)$ makes an angle $\theta$ with the $x$-axis,then the angle $\theta$ is

Two uniform spherical charge regions $S_1$ and $S_2$ having positive and negative charges overlap each other as shown in the figure. Points $O_1$ and $O_2$ are their centres and points $A, B, C, D$ and $E$ are on the line joining centres $O_1$ and $O_2$. What happens to the electric field from $C$ to $D$?

The point charges $+q, -q, -q, +q, +Q$ and $-q$ are placed at the vertices of a regular hexagon $ABCDEF$ as shown in the figure. The electric field at the centre of the hexagon '$O$' due to the five charges at $A, B, C, D$ and $F$ is twice the electric field at centre '$O$' due to charge $+Q$ at $E$ alone. The value of $Q$ is

$A$ wire of length $L = 20 \, cm$ is bent into a semicircular arc. If the two equal halves of the arc are uniformly charged with charges $+Q$ and $-Q$ respectively,where $|Q| = 10^3 \varepsilon_0$ Coulomb and $\varepsilon_0$ is the permittivity of free space,find the net electric field at the centre $O$ of the semicircular arc.

$A$ point charge $+q$ is placed at the origin. $A$ second point charge $+9q$ is placed at $(d, 0, 0)$ in a Cartesian coordinate system. The point in between them where the electric field vanishes 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 $.......\,\mu C$ $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N \cdot m^2}}{{C^2}}} \right]$