Relation between Field and Potential

Why once a choice is made regarding zero potential energy reference state, the change in potential energy is same.

The persons take different reference states for zero potential energy. The changes in potential energy for two persons is same, different, depend strictly on choice of zero of potential or become indeterminate 

What do you mean by zero potential.

Can the potential of a charged body be zero.

Derive the relation between electric field and electric potential due to a point charge.

The potential at a point $x$ due to some charge is given by equation $V\left(x\right)=\frac{20}{(x^2-4)}$ volts. Then electric field at $x=4$ is given by

The electric potential function for some electric field is defined by $V=-5x+3y+\sqrt{15}z$. The intensity of electric field (in SI unit) at point $(x, y, z)$ is

In the following graph, the magnitude of maximum electric field is :-

What is angle between electric field and equipotential surface?

The electric potential at a point $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ (all in meters) in free space is given by $\mathrm{V}=2{\mathrm{x}}^2$ Volts. Determine electric field intensity at point  $(1 \mathrm{m},2 \mathrm{m},3 \mathrm{m})$ in terms of $\hat{\mathrm{i}} \mathrm{V}/\mathrm{m}$.

For some electric field, potential at a point $(\mathrm{x}, \mathrm{y})$ is given as $\mathrm{Y}=6\mathrm{xy}+{\mathrm{y}}^2-{\mathrm{x}}^2$. Determine the electric field at this point.

Derive relation between electric potential and field.

The electric potential in complete volume of a charged conductor is same as that on its surface. Why?

If potential function is $\mathrm{V}=4\mathrm{x}+3\mathrm{y}$ volts, then calculate magnitude of electric field intensity at point $\left(2,1\right)$ meter.

The electric potential function for some electric field is defined by $\mathrm{V}=-5\mathrm{x}+3\mathrm{y}+\sqrt{15}\mathrm{z}$. The intensity of electric field $(\mathrm{in} \mathrm{S}.\mathrm{I}. \mathrm{units})$ at point $(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is

In some regions where electric field intensity $E$ is zero, the electric potential varies with distance according to

At any point $\left(\mathrm{x}, 0, 0\right)$ the electric potential $\mathrm{V}$ is $\left(\frac{1000}{\mathrm{x}}+\frac{1500}{{\mathrm{x}}^2}+\frac{500}{{\mathrm{x}}^3}\right) \mathrm{volt}$, then electric field intensity at $\mathrm{x}=1\mathrm{ }\mathrm{m}$ is

The electric potential varies in space according to the relation $\mathrm{V}=3\mathrm{x}+4\mathrm{y}$. A particle of mass $10 \mathrm{k}\mathrm{g}$ starts from rest from point $(2,3.2) \mathrm{m}$ under the influence of this field. The velocity of the particle when it crosses the $\mathrm{x}$-axis is $p\times {10}^{-3} \mathrm{m} {\mathrm{s}}^{-1}$. Find $p$. The charge on the particle is $+1 \mathrm{\mu }\mathrm{C}$. Assume $\mathrm{V}(\mathrm{x},\mathrm{y})$ are in SI units.

The potential at a point $x$ (measured in $\mathrm{\mu m}$) due to some charges situated on the x-axis is given by $V\left(x\right)=\frac{20}{\left(x^2 - 4\right)}$ volt. The electric field $E$at$\mathrm{ }x=4 \mathrm{\mu m}$ is given by

From a point charge, there is a fixed point A. At A, there is an electric field of $500 V/m$ and potential difference of 3000V. Distance between point charge and A will be