1. [tex]2.19\cdot 10^7 V[/tex]
The electric potential produced by a single point charge is
[tex]E=k\frac{q}{r^2}[/tex]
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem,
[tex]q=3.5\cdot 10^{-5} C\\r=12 cm=0.12 m[/tex]
so the electric potential is
[tex]E=(9\cdot 10^9 Nm^2 C^{-2})\frac{(3.5\cdot 10^{-5} C)}{(0.12 m)^2}=2.19\cdot 10^7 V[/tex]
2. [tex]3.5\cdot 10^6 V[/tex]
Applying the same formula used before
The electric potential produced by a single point charge is
[tex]E=k\frac{q}{r^2}[/tex]
where in this case we have
[tex]q=3.5\cdot 10^{-5} C\\r=30 cm=0.30 m[/tex]
the electric potential is
[tex]E=(9\cdot 10^9 Nm^2 C^{-2})\frac{(3.5\cdot 10^{-5} C)}{(0.30 m)^2}=3.5\cdot 10^6 V[/tex]
3. 644 J
The work done to carry a charge q through a potential difference of [tex]\Delta V[/tex] is given by
[tex]W=q\Delta V[/tex]
In this problem, we have
[tex]q=3.5\cdot 10^{-5}C[/tex] is the charge
[tex]\Delta V=V(12 cm)-V(30 cm)=2.19\cdot 10^7 V-3.5\cdot 10^6 V=1.84\cdot 10^7 V[/tex] is the potential difference
Therefore, the work done is
[tex]W=(3.5\cdot 10^{-5} C)(1.84\cdot 10^7 V)=644 J[/tex]
