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A parallel plate capacitor filled with mica with [tex]\varepsilon_r = 5[/tex] is connected to a 10 V battery. The area of the parallel plate is [tex]6 \, \text{m}^2[/tex] and the separation distance is [tex]6 \, \text{mm}[/tex].

(a) Find the capacitance and stored charge.

(b) After the capacitor is fully charged, the battery is disconnected and the dielectric is removed carefully. Calculate the new values of capacitance, stored energy, and charge.

Answer :

Final answer:

The capacitance of the parallel plate capacitor is 5000 F and the stored charge is 50000 C. After the capacitor is fully charged and the dielectric is removed, the new values are: capacitance = ε0A/d, stored energy = 0.5 * CV2, and charge = CV.

Explanation:

To find the capacitance and stored charge of the parallel plate capacitor:

  1. Given: ε = 5, V = 10 V, A = 6 m2, d = 6 mm = 0.006 m
  2. Capacitance (C) can be calculated using the formula C = εA/d.
  3. Substituting the given values, C = 5 * 6 / 0.006 = 5000 F
  4. Stored charge (Q) can be calculated using the formula Q = CV.
  5. Substituting the values, Q = 5000 F * 10 V = 50000 C

After the capacitor is fully charged and the dielectric is removed:

  1. The new values of capacitance, stored energy, and charge can be calculated using the same formulas.
  2. Since the dielectric is removed, the permittivity (ε) becomes the permittivity of free space (ε0).
  3. Capacitance (C) can be calculated using the formula C = ε0A/d.
  4. Stored energy (U) can be calculated using the formula U = 0.5 * CV2.
  5. Charge (Q) can be calculated using the formula Q = CV.

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