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An isolated 16 μF parallel plate air capacitor has a potential difference of 1000 V (Figure a). A dielectric slab having relative permittivity (i.e. dielectric constant) = 5 is introduced to fill the space between the two plates completely (Figure b)Calculate :(i) the new capacitance of the capacitor.(ii) the new potential difference between the two plates of the capacitor. |
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Answer» Here C0 = 16μF, K = 5 (i) When the dielectric of dielectric constant K is introduced between the plates of the capacitor, the capacity increases K times, new capacity = KC0 = 5 × 16 = 80μF, v0 = 1000v (ii) The new potential difference V is given by V = \(\frac{V_0}{K}\,= \frac{1000}{5}\) = 200V |
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