Respuesta :
1) The electric potential at a distance r from a single point charge is given by
[tex]V(r) = k \frac{q}{r} [/tex]
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.
The charge in this problem is
[tex]q=2.2 \mu C =2.2 \cdot 10^{-6} C[/tex]
So the potential at distance [tex]r=6.3 m[/tex] is
[tex]V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V [/tex]
2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
[tex]V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V [/tex]
[tex]V(r) = k \frac{q}{r} [/tex]
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.
The charge in this problem is
[tex]q=2.2 \mu C =2.2 \cdot 10^{-6} C[/tex]
So the potential at distance [tex]r=6.3 m[/tex] is
[tex]V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2}) \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V [/tex]
2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
[tex]V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V [/tex]