Respuesta :
Answer:
[tex] E_f = \frac{V}{\frac{d}{2}}= 2 \frac{V}{d}= 2* 6x10^4 \frac{N}{C} = 1.2x10^5 \frac{N}{C}[/tex]
Explanation:
For this case we know that the electric field is given by:
[tex] E= 6 x 10^4 \frac{N}{C}[/tex]
And we want to find the final electric field assuming that the separation is halved and becomes d/2.
For this case we can use the following two equations:
[tex] C = \epsilon_o \frac{A}{d} = \frac{Q}{V}[/tex] (1)
[tex] E = \frac{\sigma}{\epsilon_o}[/tex] (2)
Where Q represent the charge, V the voltage, d the distance, A the area.
We can rewrite the equation (2) like this:
[tex] E = \frac{\sigma}{\epsilon_o} = \frac{Q}{A \epsilon_o}[/tex] (3)
And we can solve for Q from equation (1)like this:
[tex] Q = \frac{\epsilon_o A V}{d}[/tex]
And if we replace into equation (3) the previous result we got:
[tex] E = \frac{\epsilon_o A V}{A d \epsilon_o} = \frac{V}{d}[/tex]
And since the the electric field not change and the distance would be the half we have that the final electric field would be given by:
[tex] E_f = \frac{V}{\frac{d}{2}}= 2 \frac{V}{d}= 2* 6x10^4 \frac{N}{C} = 1.2x10^5 \frac{N}{C}[/tex]
