Question

The electric field strength 5.0 cm from a very long charged wire is 3300 N/C .What is the electric field strength 10.0 cm from the wire?

Answer 1

To solve this problem we will apply the concepts related to the electric field. The relationship between state two and state 1 will be carried out on said electric field, and from there we will proceed to replace.

The electric field due to a wire a distance $r_1$ is,

$E_1 = \frac{\lambda}{2\pi \epsilon_0 r_1}$

The electric field due to a wire a distance $r_2$ is,

$E_2 = \frac{\lambda}{2\pi \epsilon_0 r_2}$

The ratio is

$\frac{E_2}{E_1} = \frac{\frac{\lambda}{2\pi \epsilon_0 r_2}}{\frac{\lambda}{2\pi \epsilon_0 r_1}}$

The electric field strength is,

$E_2 = E_1 (\frac{\frac{\lambda}{2\pi \epsilon_0 r_2}}{\frac{\lambda}{2\pi \epsilon_0 r_1}})$

$E_2 = E_1 (\frac{r_1}{r_2})$

$E_2 = (3300N/C)(\frac{5cm}{10cm})$

$E_2 = 1650N/C$

Therefore the electric field strenth 10cm from the wire is 1650N/C

Answer 2

The electric field strength 10 cm from the wire is 1650 N/C.

The electric field is inversely proportional to distance from charge.

Given that, The electric field strength 5.0 cm from a very long charged wire is 3300 N/C.

    $E_{1}=3300N/C, r_{1}=5cm,r_{2}=10cm$

Relation is,

                     $\frac{E_{1}}{E_{2}}=\frac{r_{2}}{r_{1}} \\ \\ \frac{3300}{E_{2}}=\frac{10}{5}\\ \\ E_{2}=\frac{3300}{2}=1650N/C$

Hence, the electric field strength 10 cm from the wire is 1650 N/C.

Learn more:

https://brainly.com/question/11509296