Three charges $q, \sqrt 2q, 2q$ are placed at the corners $A, B$ and $C$ respectively of the square $ABCD$ of side $'a'$ then potential at point $'D'$

Value of potential at a point due to a point charge is

The electric potential $V(x, y, z)$ for a planar charge distribution is given by: 

$V\left( {x,y,z} \right) = \left\{ {\begin{array}{*{20}{c}}
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, < \, - d}\\
{ - {V_0}{{\left( {1 + \frac{x}{d}} \right)}^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\, - \,d\, \le x < 0}\\
{ - {V_0}\left( {1 + 2\frac{x}{d}} \right)\,\,\,\,\,\,\,\,\,\,\,for\,0\, \le x < d}\\
{ - 3{V_0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,x\, > \,d}
\end{array}} \right.$

where $-V_0$ is the potential at the origin and $d$ is a distance. Graph of electric field as a function of position is given as

Charge is uniformly distributed on the surface of a hollow hemisphere. Let $O$ and $A$ be two points on the base of the hemisphere and $V_0$ and $V_A$ be the electric potentials at $O$ and $A$ respectively. Then,

Two large vertical and parallel metal plates having a separation of $1 \ cm$ are connected to a $DC$ voltage source of potential difference $X$. A proton is released at rest midway between the two plates. It is found to move at $45^{\circ}$ to the vertical $JUST$ after release. Then $X$ is nearly

A conductor with a positive charge