Potential in the $x-y$ plane is given as $V = 5(x^2 + xy) \, V$. The electric field at the point $(1, -2)$ will be

Electric field in a region is given by $\vec{E} = Ax\hat{i} + By\hat{j}$, where $A = 10 \ V/m^2$ and $B = 5 \ V/m^2$. If the electric potential at a point $(10, 20)$ is $500 \ V$, then the electric potential at the origin is . . . . . . $V$.

Assume that an electric field $E=20 x^2 \hat{i}$ exists in space. If $V_0$ is the potential at the origin and $V_A$ is the potential at $x=3 \ m$,then the potential difference $V_A-V_0$ in volts is

Electric field at a point $(x, y, z)$ is represented by $\vec E = 2x\hat i + y^2\hat j$. If the potential at $(0, 0, 0)$ is $2 \, V$,find the potential at $(1, 1, 1)$. (in $/3$)

The variation of potential with distance $R$ from a fixed point is as shown in the graph. The electric field at $R = 5\,m$ is......$V/m$.

In a region,the electric field is $(30 \hat{i} + 40 \hat{j}) \text{ NC}^{-1}$. If the electric potential at the origin is zero,the electric potential at the point $(1 \text{ m}, 2 \text{ m})$ is