To find the dielectric constant of the material inserted between the plates of the capacitor, you can use the relationship between the capacitance with the dielectric material and the capacitance with air. Here's the detailed, step-by-step solution:
1. Identify the given values:
- The initial capacitance with air as the dielectric: [tex]\( C_{\text{air}} = 3 \, \mu F \)[/tex]
- The capacitance with the dielectric material: [tex]\( C_{\text{dielectric}} = 15 \, \mu F \)[/tex]
2. Understand the relationship:
The dielectric constant (commonly denoted as [tex]\( \kappa \)[/tex] or [tex]\( k \)[/tex]) is the factor by which the capacitance increases when a dielectric material is inserted. The capacitance with the dielectric material is given by:
[tex]\[
C_{\text{dielectric}} = k \cdot C_{\text{air}}
\][/tex]
where [tex]\( k \)[/tex] is the dielectric constant.
3. Set up the equation using the given values:
Substitute the given values of [tex]\( C_{\text{air}} \)[/tex] and [tex]\( C_{\text{dielectric}} \)[/tex] into the equation:
[tex]\[
15 \, \mu F = k \cdot 3 \, \mu F
\][/tex]
4. Solve for the dielectric constant [tex]\( k \)[/tex]:
Divide both sides of the equation by [tex]\( 3 \, \mu F \)[/tex] to isolate [tex]\( k \)[/tex]:
[tex]\[
k = \frac{15 \, \mu F}{3 \, \mu F}
\][/tex]
5. Calculate the result:
[tex]\[
k = 5
\][/tex]
Thus, the dielectric constant of the material is [tex]\( 5 \)[/tex].
