Figure shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and lengths l1 and l2. The left half of the dielectric slab has a dielectric constant K1 and the right half K2. Neglecting any friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium

F=ε0bV2(K−1)2d⇒V∝1k−1Ratio of the EMFs will be K2−1K1−1

Figure shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and lengths l1 and l2. The left half of the dielectric slab has a dielectric constant K1 and the right half K2. Neglecting any friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium

F=ε0bV2(K−1)2d⇒V∝1k−1Ratio of the EMFs will be K2−1K1−1

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Figure shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and lengths l₁ and l₂. The left half of the dielectric slab has a dielectric constant K₁ and the right half K₂. Neglecting any friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium

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$\sqrt{\frac{K_{2} - 1}{K_{1} - 1}}$

$\sqrt{\frac{K_{1} - 1}{K_{2} - 1}}$

$\sqrt{\frac{K_{1} + 1}{K_{2} + 1}}$

$\sqrt{\frac{K_{2} + 1}{K_{1} + 1}}$

answer is A.

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Detailed Solution

$F = \frac{ε_{0} b V^{2} (K - 1)}{2 d}$

$\Rightarrow V \propto \frac{1}{\sqrt{k - 1}}$

Ratio of the EMFs will be $\sqrt{\frac{K_{2} - 1}{K_{1} - 1}}$

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