Do free electrons travel to region of higher potential or lower potential ?
No. There is negative charge on free electrons. The free electrons experience electrostatic force in a direction opposite to the direction of electric field. The direction of electric field is always from higher potential to lower. Hence, free electrons will move from lower potential to higher potential.
Two identical metal balls of radius $r$ are at a distance $a (a >> r)$ from each other and are charged, one with potential $V_1$ and other with potential $V_2$. The charges $q_1$ and $q_2$ on these balls in $CGS$ esu are
The give graph shown variation (with distance $r$ from centre) of
$64$ identical drops each charged upto potential of $10\,mV$ are combined to form a bigger dorp. The potential of the bigger drop will be $..........\,mV$
Consider a sphere of radius $R$ with uniform charge density and total charge $Q$. The electrostatic potential distribution inside the sphere is given by $\theta_{(r)}=\frac{Q}{4 \pi \varepsilon_{0} R}\left(a+b(r / R)^{C}\right)$. Note that the zero of potential is at infinity. The values of $(a, b, c)$ are
Six charges are placed around a regular hexagon of side length a as shown in the figure. Five of them have charge $q$, and the remaining one has charge $x$. The perpendicular from each charge to the nearest hexagon side passes through the center $O$ of the hexagon and is bisected by the side.
Which of the following statement($s$) is(are) correct in SI units?
$(A)$ When $x=q$, the magnitude of the electric field at $O$ is zero.
$(B)$ When $x=-q$, the magnitude of the electric field at $O$ is $\frac{q}{6 \pi \epsilon_0 a^2}$.
$(C)$ When $x=2 q$, the potential at $O$ is $\frac{7 q}{4 \sqrt{3} \pi \epsilon_0 a}$.
$(D)$ When $x=-3 q$, the potential at $O$ is $\frac{3 q}{4 \sqrt{3} \pi \epsilon_0 a}$.