$A$ 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?

Two equal negative charges $-q$ are fixed at points $(0, -a)$ and $(0, a)$ in the $x-y$ plane. $A$ positive charge $Q$ is released from rest at a point $(2a, 0)$. The charge $Q$ will

At a certain distance from a point charge,the field intensity is $500 \, Vm^{-1}$ and the potential is $-3000 \, V$. The distance to the charge and the magnitude of the charge respectively are

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:

Three charges $2q, -q, -q$ are located at the vertices of an equilateral triangle. At the centre of the triangle:

Four charges are placed on the corners of a square as shown in the figure,with a side length of $5 \, cm$. If $Q = 1 \, \mu C$,then the electric field intensity at the centre will be: