The point charges $Q$ and $-2Q$ are placed at a distance $r$ apart. If the electric field at the location of $Q$ (point $A$) due to $-2Q$ is $\vec E$,then the electric field at the location of $-2Q$ (point $B$) due to $Q$ will be:

$A$ spherical conductor of radius $2 \text{ cm}$ is uniformly charged with $3 \text{ nC}$. What is the electric field at a distance of $3 \text{ cm}$ from the center of the sphere?

The electric field at the center of a semi-circle of radius $a$ and linear charge density $\lambda$ is given by:

Total charge $-Q$ is uniformly spread along the length of a ring of radius $R$. $A$ small test charge $+q$ of mass $m$ is kept at the centre of the ring and is given a gentle push along the axis of the ring.
$(a)$ Show that the particle executes a simple harmonic oscillation.
$(b)$ Obtain its time period.

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 thrice the electric field at the centre '$O$' due to charge $+Q$ at $E$ alone. The value of $Q$ is:

Two point electric charges $+10^{-8} \text{ C}$ and $-10^{-8} \text{ C}$ are placed $0.1 \text{ m}$ apart. Find the magnitude of the total electric field at the centre of the line joining the two charges.