Answer:
5.8
Explanation:
- The capacitance of any capacitor, by definition, is as follows:
[tex]C =\frac{Q}{V}[/tex]
- For a parallel-plate capacitor, applying Gauss'Law and the definition of electric potential, it can be found that the capacitance can be expressed as follows:
[tex]C =\frac{\epsilon*A}{d}[/tex]
- Where ε = ε₀*εr (being εr the dielectric constant of the material that fills the space between the plates, A the area of one of the plates, and d, the separation between them).
- If the capacitor is disconnected from the battery, due to the conservation of the charge, the charge can't change after being disconnected, so Qf = Q₀.
- If we apply the definition of capacitance, we can find the relationship between the initial and the final capacitance, as follows:
[tex]C_{0} = \frac{Q_0}{V_{0} } \\ \\ C_{f} = \frac{Q_f}{V_{f} } \\ \\ Q_{0} = Q_{f} = C_{0} * V_{0} = C_{f} * V_{f} \\ \\ \frac{C_{0} }{C_{f} } = \frac{V_{f}}{V_{0}} = \frac{525 V}{90V} = 5.8[/tex]
- If we look at the expression for the capacitance in a parallel-plate capacitor, if A and d remain constant, the only parameter changing is the dielectric constant ε, so we can write the following equation:
[tex]\frac{C_{0}}{C_{f} } = \frac{\epsilon r*\epsilon 0*A}{d} *\frac{d}{\epsilon 0*A} =5.8[/tex]
⇒εr = 5.8
- The dielectric constant of the material that was initially used to fill the gap between the plates is 5.8.
