In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric slabs of thickness 5 cm each and dielectric constant $K_{1} = 2$ and $K_{2} = 4$ respectively, are inserted between the plates. A potential difference of 100V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is

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$\frac{2000}{3} ε_{\begin{array}{l} 0 \end{array}}$

$\frac{- 2000}{3} ε_{0}$

$- 250 ε_{0}$

$\frac{- 1000}{3} ε_{0}$

answer is A.

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Detailed Solution

$V_{1} : V_{2} = \frac{1}{C_{1}} : \frac{1}{C_{2}} = 2 : 1 V_{1} + V_{2} = 100 V ∴ V_{1} = \frac{200 V}{3}$
charge on plates $= Q = \frac{2 ε_{0} A}{d} V_{1} = \frac{400}{3} \frac{ε_{0} A}{d}$
$∴ σ = \frac{400 ε_{0}}{3 D} = \frac{400 ε_{0}}{3 \times 5 \times 10^{- 2}} = \frac{8000 ε_{0}}{3}$
Net bound charge
Density at interface
$= + σ (1 - \frac{1}{K_{1}}) - σ (1 - \frac{1}{K_{2}}) = + σ (1 - \frac{1}{2}) - σ (1 - \frac{1}{4}) = + \frac{σ}{2} - \frac{3 σ}{4} = \frac{- σ}{4} = \frac{- 2000 ε_{0}}{3}$