A potential difference of 200 volts is applied across the two plates of a parallel plate capacitor. The area of each plate is 100π cm^2 and separation between the plates is 1 mm. The space between is filled with a mica having k=6. Calculate (i) Charge on each plate and (ii) Electric field intensity within the sheets.
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- A 4 μF parallel-plate capacitor with a dielectric has a charge of 42 μC on each plate. The distance between the plates is 0.007 m and the dielectric constant is 153. Determine the area of each plate and the E-field between the plates. A = E =A parallel-plate capacitor has a plate area of 0.2m² and a plate separation of 0.1 mm. To obtain an electric field of 2.0 x 10⁶ V/m between the plates, the magnitude of the charge on each plate should be: a. 9.0 x 10⁷ Cb. 2.0 x 10⁻⁶ Cc. 4.0 x 10⁻⁶ Cd. 7.1 x 10⁻⁶ CA parallel plate capacitor has capacitance C0 = 15.0 pF when there is air between the plates. The separation between the plates is 1.00 mm. A dielectic with K=3.50 is inserted between the plates of the capacitor, completely feeling the volume between the plates. The capacitor is connected with a battery. Find the magnitude of charge on each plate if the electric field between the plates is 3.00 x 104 V/m.
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