Answer:
The value is
![E = 14.12 \ N/C](https://img.qammunity.org/2021/formulas/physics/college/lrmqoqss1z6ktwjrugygtnfxlpoxvw8awf.png)
Step-by-step explanation:
From the question we are told that
The radius of the cylinder is
![r = 12 \ cm = 0.12 \ m](https://img.qammunity.org/2021/formulas/physics/college/esfvzqk7zcjak3ir1jtr8kfxrjd7j3g3tr.png)
The density of the charge is
![\rho = 5.0 \ nC/m^3 = 5.0*10^(-9) \ C/m^3](https://img.qammunity.org/2021/formulas/physics/college/k53kezmqdjomxlo7s7o1kxr2zb15i2so8h.png)
The position consider is a = 5.0 cm = 0.05 m
Gnerally from the magnitude of the magnetic field is mathematically represented as
![E = (\rho * s)/( 2 * \epsilon_o )](https://img.qammunity.org/2021/formulas/physics/college/uu6vnzi8j6iikqfwr1l8os9o357j4r6gq1.png)
Here
is the permittivity of free space with value
![\epsilon_o = 8.85 *10^(-12) \ C/(V \cdot m)](https://img.qammunity.org/2021/formulas/physics/college/ret5htsnip9r242gwn2c8lumvstue7yooc.png)
=>
![E = (5.0*10^(-9) * 0.05)/( 2 * 8.85*10^(-12) )](https://img.qammunity.org/2021/formulas/physics/college/fjbc4rwldiw5ymsntclbxe1tpuwjqrkndr.png)
=>
![E = 14.12 \ N/C](https://img.qammunity.org/2021/formulas/physics/college/lrmqoqss1z6ktwjrugygtnfxlpoxvw8awf.png)