As shown in the figure,charges $+q$ and $-q$ are placed at the vertices $B$ and $C$ of an isosceles triangle. The potential at the vertex $A$ is

The ratio of electric potentials at the point $E$ to that at the point $F$ is

The side length of an equilateral triangle is $d$. Three charges,each of magnitude $Q$,are placed at the three vertices. If $P$ is the midpoint of one side,find the electric potential $V_P$ at point $P$.

Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+Q$. $A$ point charge $-q$ $(q \ll Q)$ with mass $m$ is released from rest at a distance of $3R$ from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is:
($\epsilon_0$ is the permittivity of vacuum, neglect gravitational forces).

Assertion: Electrons move away from a region of higher potential to a region of lower potential.
Reason: An electron has a negative charge.

$A$ hollow metal sphere of radius $15 \,cm$ is charged such that the potential on its surface is $20 \,V$. Then, the potential at the centre of the sphere is (in $V$)