A uniform electric field having a magnitude ${E_0}$ and direction along the positive $X - $ axis exists. If the potential $V$ is zero at $x = 0$, then its value at $X = + x$ will be
${V_{x}} = + x{E_0}$
${V_x} = - x{E_0}$
${V_x} = + {x^2}{E_0}$
${V_x} = - {x^2}{E_0}$
The potential $V$ is varying with $x$ and $y$ as $V\, = \,\frac{1}{2}\,\left( {{y^2} - 4x} \right)\,volt.$ The field at ($1\,m, 1\,m$ ) is
The potential gradient is a
In a certain region of space, the potential is given by : $V = k[2x^2 - y^2 + z^2].$ The electric field at the point $(1, 1, 1) $ has magnitude =
If potential (in volts) in a region is expressed as $V (x,y,z) =6xy-y+2yz $ the electric field (in $N/C$) at point $(1, 1, 0)$ is
The variation of potential with distance $x$ from a fixed point is as shown in figure. The electric field at $x =13\,m$ is......$volt/meter$