A parallel plate capacitor of capacitance 100 mu F Is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes

A parallel plate capacitor of capacitance 100 mu F Is charged to 500 V

1.	A parallel plate capacitor of capacitance 100 mu F Is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes
Answer» 250 V 500 V 1000 V 2000 V Solution :Here, ` "" C.= 2C`, since the charge REMAINS the same , `q = C.V. = CV RARR V. = (CV)/(2C) = (500)/(2) = 250 `C

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A parallel plate capacitor of capacitance 100 mu F Is charged to 500 V. The plate separation is then reduce to half its original value. Then the potential on the capacitor becomes

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