A battery sets up an electric field of $25 \mathrm{N} {\mathrm{C}}^{-1}$ inside a uniform wire of length $2 \mathrm{m}$ and resistance of $5 \mathrm{\Omega }$. Find the current through the wire.

The figure shows the current $I$ in a single-loop circuit with a battery $B$ and resistance $R$ (and wires of negligible resistance). Then find the order of following at the point $a\mathit{,}\mathit{ }b$ and $c$.

(a) The magnitude of the current,

(b) The electric potential, and

(c) The electric potential energy of the charge carriers (electron), the greatest first.

The magnitude of electric force on $2 \mathrm{\mu C}$ charge placed at the centre $O$ of two equilateral triangles each of side $10 \mathrm{cm}$, as shown in the figure is $P$. If charge $A\mathit{,}\mathit{ }B\mathit{,}\mathit{ }C\mathit{,}\mathit{ }D\mathit{,}\mathit{ }E$ and $F$ are $2 \mathrm{\mu C}, 2 \mathrm{\mu C}, 2 \mathrm{\mu C}, -2 \mathrm{\mu C}, – 2 \mathrm{\mu C}, – 2 \mathrm{\mu C}$ respectively, then $P$ is:

Two-point charges of the same magnitude and opposite sign are fixed at points $A$ and $B$. A third small point charge is to be balanced at point $P$ by the electrostatic force due to these two charges. The point $P$

Two-point charges $a$ and $b,$ whose magnitudes are the same are positioned at a certain distance from each other with $a$ at the origin. The graph is drawn between electric field strength at points between $a$ and $b$ and distance $x$ from $a$. $E$ is taken positive if it is along the line joining from $a$ to $b$. From the graph, it can be decided that

A charged particle $‘q’$ is shot from a large distance with speed $v$ towards a fixed charged particle $Q$. It approaches $Q$ upto the closest distance $r$ and then returns. If $q$ were given a speed $‘2v’$, the closest distance of approach would be: