A parallel plate capacitor of capacitance 5 μ F and plate separation 6 cm is connected to a 1 V battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is

Question

A parallel plate capacitor of capacitance 5 μ F and plate separation 6 cm is connected to a 1 V battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is (A) 2 μ C (B) 3 μ C (C) 5 μ C (D) 10 μ C

Tardigrade Admin · Accepted Answer

∴ Charge on capacitor plates now will be Q' = CV = 10μ F × 1V = 10 μ C Additional charge transferred = Q' - Q = 10 μ C = 5 μ C

$2 μ C$

$3 μ C$

$5 μ C$

$10 μ C$

$∴$ Charge on capacitor plates now will be
$Q^{'} = C V = 10 μ F \times 1 V = 10 μ C$
Additional charge transferred = $Q^{'} - Q = 10 μ C$
= $5 μ C$

Q. A parallel plate capacitor of capacitance 5μF and plate separation 6cm is connected to a 1V battery and charged. A dielectric of dielectric constant 4 and thickness 4cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is

2μC

3μC

5μC

10μC

∴ Charge on capacitor plates now will be Q′=CV=10μF×1V=10μC Additional charge transferred = Q′−Q=10μC = 5μC

$2 μ C$

$3 μ C$

$5 μ C$

$10 μ C$

$∴$ Charge on capacitor plates now will be
$Q^{'} = C V = 10 μ F \times 1 V = 10 μ C$
Additional charge transferred = $Q^{'} - Q = 10 μ C$
= $5 μ C$