Respuesta :
Answer:
Correct -> E(0) = 0
Correct -> [tex]\lim_{r \to \infty} E(r) = 0[/tex]
Correct -> The maximum electric field occurs when r = rb.
Explanation:
The electric field formula due to a particle in space is
[tex]\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{q}{r^2}\^{r}[/tex]
However, inside the ball the electric field is different.
Applying Gauss' Law inside the ball gives:
[tex]\int {\vec{E}} \, da = \frac{Q_{\rm enc}}{\epsilon_0}\\ E4\pi r^2 = \frac{qr^3}{(rb)^3\epsilon_0}\\E = \frac{1}{4\pi\epsilon_0}\frac{r}{(rb)^3}[/tex]
- At r = 0, the above formula yields 0 as well. Therefore the statement E(0) = 0 is correct.
- Due to the above formula,
[tex]E(rb) = \frac{1}{4\pi \epsilon_0}\frac{q(rb)}{(rb)^3} = \frac{1}{4\pi\epsilon_0}\frac{q}{(rb)^2} \neq 0[/tex]
Therefore, the statement "E(rb) = 0" is wrong.
- [tex]\lim_{r \to \infty} E(r) = 0[/tex]
This statement is correct. Since r is in the denominator, 1/∞ can be taken zero.
- The maximum electric field does not occur at r = 0.
- If 'rb' is the radius of the ball, then the maximum electric field occurs when r = rb.
- At infinity, the electric field goes to zero, therefore the maximum electric field does not occur as r -> infinity.
