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A parallel plate air capacitor is connected to a battery. Its charge, potential difference, electric field and the stored energy between the plates are Q_0,V_0,E_0" and "U_0 respectively. Keeping the battery connection uncharged, the capacitor is completely filled with a dielectric material. The charge, potential difference, electric field and energy stored become Q,V,E and U respectively. Which of the following is correct? Q gt Q_0. |
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Answer» Solution :When the dielectric material is inserted in the capacitor, its CAPACITANCE INCREASES. As the capacitor is still CONNECTED to the BATTERY, so its voltage will remain CONSTANT. Since, charge = capacitance `xx` voltage, so `Qgt Q_0`. |
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