Two large plane parallel conducting plates are kept $10 \ cm$ apart as shown in the figure. The potential difference between them is $V$. The potential difference between the points $A$ and $B$ (shown in the figure) is

Two points $A$ and $B$ on the $Y$-axis are at distances of $12.3 \ cm$ and $12.5 \ cm$ from the origin,respectively. The potentials at these points are $56 \ V$ and $54.8 \ V$,respectively. What is the component of the force on a charge of $4 \ \mu C$ placed at point $A$ on the $Y$-axis?

The electric potential $V$ is given as a function of distance $x$ $(m)$ by $V = (2x^2 + 10x - 9) \text{ V}$. The value of the electric field at $x = 1 \text{ m}$ is: (in $\text{ V/m}$)

The potential $\phi(x, y)$ of an electrostatic field $\vec{E} = a(y \hat{i} + x \hat{j})$ is [where $a$ is a constant and $\hat{i}$ and $\hat{j}$ are unit vectors along $X$ and $Y$ axes].

Electric potential in a region is varying according to the relation $V = \frac{3x^2}{2} - \frac{y^2}{4}$,where $x$ and $y$ are in $m$ and $V$ is in $V$. Electric field intensity (in $N/C$) at a point $(1 \, m, 2 \, m)$ is:

For the given figure,the value of the electric field is ......... $V/m$.