The ratio of charge q 2 q 1 on the lower plate of capacitor is :-

Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitor236Medium2 minQB

PhysicsMediumsingle choice

Passage / Comprehension

Figure shows a parallel plate capacitor having an insulator of dielectric constant $K = 2$ . The plate area of the capacitor is A & separation between them is ' d '; while the surface area of the dielectric is $A /3$ and its height is $d /3$ . The capacitor is connected to a battery of potential difference 'V'. Let $q_{1}$ be the free charge that appears at the upper surface of lower plate, just below the dielectric, while $q_{2}$ is the surface charge that appears on the other side as shown.\\

The ratio of charge $\frac{q _{1}}{q _{2}}$ on the lower plate of capacitor is :-

Options:

Answer:

Solution:

Same potential difference across both branches, so charges are in ratio of branch capacitances

$\frac{q _{1}}{q _{2}} = \frac{3}{5}$

Answer (B)

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:236

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...