Question

PLEASE HELP

Answer 1

(a) the electric field generated by a single point charge is radial, and its direction is determined by the sign of the source charge. In particular, when the source charge is positive as in this example, the field lines are directed away from the charge (see attached figure).

(b) The electric field strength at distance r from the source charge is given by
$E=k_e \frac{Q}{r^2}$
where ke is the Coulomb's constant and Q is the source charge. In our problem, the distance is r=2.00 cm=0.02 m, so the electric field at that point is
$E=(8.99 \cdot 10^9 N m^2 C^{-2}) \frac{2.0 \cdot 10^{-5}C}{(0.02 m)^2}=4.5 \cdot 10^8 N/C$

The force experienced by the test charge in this electric field is equal to
$F=qE$
where $q=+5 \mu C=+5 \cdot 10^{-6}C$ is our test charge. Substituting, we find that the force is equal to
$F=qE=(5\cdot 10^{-6}C)(4.5 \cdot 10^8 N/C)=2250 N$

(c) The electric potential at a distance r from the source (positive) charge is given by
$V=k_e \frac{Q}{r}$
and the electric potential energy of a test charge q at distance r from the source charge is
$U=qV=k_e \frac{qQ}{r}$
We can see that when the test charge q is moved closer to the source Q, the distance r decreases, and so the potential energy U of the charge increases. So, the change in potential energy is positive.