Respuesta :
Answer:
so the distance between two points are
[tex]d = 0.246 \times 10^{-3} m[/tex]
Explanation:
Surface charge density of the charged plane is given as
[tex]\sigma = 7.2 \mu C/m^2[/tex]
now we have electric field due to charged planed is given as
[tex]E = \frac{\sigma}{2\epsilon_0}[/tex]
now we have
[tex]E = \frac{7.2 \times 10^{-6}}{2(8.85 \times 10^{-12})}[/tex]
[tex]E = 4.07 \times 10^5 N/C[/tex]
now for the potential difference of 100 Volts we can have the relation as
[tex]E.d = \Delta V[/tex]
[tex]4.07 \times 10^5 (d) = 100[/tex]
[tex]d = \frac{100}{4.07 \times 10^5}[/tex]
[tex]d = 0.246 \times 10^{-3} m[/tex]
