A parallel plate capacitor with plates separated by air acquires 1 $μ$ C of charge when connected to a battery of 500V. The plates still connected to the battery are then immersed in benzene ( $k = 2.25$ ). Then a charge flows from the battery is :

$1.25 μ$ C
$2.28 μ$ C
$1 / 4 μ$ C
$4.56 μ$ C

Question

A parallel plate capacitor with plates separated by air acquires 1 $μ$ C of charge when connected to a battery of 500V. The plates still connected to the battery are then immersed in benzene ( $k = 2.25$ ). Then a charge flows from the battery is :

$1.25 μ$ C
$2.28 μ$ C
$1 / 4 μ$ C
$4.56 μ$ C

Toppr Admin · Accepted Answer

Initial charge Q1-1-x3BC-cV1-500V-As the plates are still connected voltage is constant-x2234- when plates are dipped in benzene capacitance increases by k-k-x3B5-0Ad-x2234- As voltage is constant Q also increases by k-x2234-Q2-kQ1charge flown from battery-kQ1-x2212-Q11-x3BC-c-2-x22C5-25-x2212-1-1-xD7-10-x2212-6-1-x22C5-25-1-x22C5-25-x3BC-c-

A parallel plate capacitor with plates separated by air acquires 1 μC of charge when connected to a battery of 500V. The plates still connected to the battery are then immersed in benzene (k=2.25). Then a charge flows from the battery is :1.25μ C2.28μ C1/4μ C4.56μ C

Initial charge Q1=1μcV1=500V.As the plates are still connected voltage is constant.∴ when plates are dipped in benzene capacitance increases by k=(kε0Ad)∴ As voltage is constant Q also increases by k.∴Q2=kQ1charge flown from battery=kQ1−Q11μc(2⋅25−1)1×10−6(1⋅25)=1⋅25μc.

A parallel plate capacitor with plates separated by air acquires 1 $μ$ C of charge when connected to a battery of 500V. The plates still connected to the battery are then immersed in benzene ( $k = 2.25$ ). Then a charge flows from the battery is :

$1.25 μ$ C
$2.28 μ$ C
$1 / 4 μ$ C
$4.56 μ$ C