If the electric potential is given by $V = (5x^2 + 10x - 9) \ V$,what is the electric field at $x = 1 \ m$ in $V/m$?

Write the relation between electric field and electrostatic potential.

The figure shows a set of equipotential surfaces. The magnitude and direction of the electric field that exists in the region is .........

Let $V$ and $E$ be the potential and electric field intensity at a point,respectively. Then:

Two parallel plates are separated by a distance of $5 \, mm$. The potential difference between them is $50 \, V$. $A$ particle of mass $10^{-15} \, kg$ and charge $10^{-11} \, C$ enters the region with a velocity of $10^7 \, m/s$. The acceleration of the particle will be:

The potential (in volts) of a charge distribution is given by $V(z) = 30 - 5z^2$ for $|z| \le 1 \ m$ and $V(z) = 35 - 10|z|$ for $|z| \ge 1 \ m$. $V(z)$ does not depend on $x$ and $y$. If this potential is generated by a constant charge per unit volume $\rho_0$ (in units of $\varepsilon_0$) which is spread over a certain region,then choose the correct statement.