It is found that air breaks down electrically, when the electric field is $ 3 \times 10^{6} \mathrm{V} / \mathrm{m} . $ What is the potential to which a sphere of radius $1 \mathrm{m} $ can be raised, before sparking takes place?
In moving from A to B along an electric field line, the wok done by the electric field on an electron is $6.4 \times 10^{-19}$ J. If $\phi _1$ and $\phi _2$ are equipotential surfaces, then the potential difference $V _b-V _A $ is
The equation of an equipotential line is an electric field is y = 2x, then the electric field strength vector at (1, 2) may be
The electric potential in a certain region along the x-axis varies with x according to the relation $V(x) = 5 - 4x^2$. Then, the correct statement is :
A point charge q moves from point P to a point S along a path PQRS in a uniform electric field E pointing parallel to the x-axis. The coordinates of P, Q. R and S are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$. The work done by the field in the above process is :
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 =
A charge of 3C moving in a uniform electric field experiences a force of $3000 N$. The potential difference between two points situated in the field at a distance $1 cm$ from each other will be
The potential at a point $x$ (measured in $\mu m )$ due to somecharges situated on the $x$ -axis is given by $V ( x ) = 20 / \left( x ^ { 2 } - 4 \right)$Volts. The electric field $E$ at $x = 4 \mu m$ is given by
Variation in potential is maximum if one goes :
The electric field lines are closer together near object $A$ than they are near object $B$. We can conclude that :