A parallel plate capacitor is made of two plates of length l, width w and separated by distance d. A dielectric slab (dielectric constant K) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force F = -(∂ U)/(∂ x) where U is the energy of the capacitor when dielectric is inside the capacitor up to distance x (see figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is: (a) (Q² d)/(2wl²ε₀)K (b) (Q² w)/(2dl²ε₀) (K - 1) (c) (Q² d)/(2wl²ε₀) (K - 1) (d) (Q² w)/(2dl²ε₀) K

Answer: (c). With slab in up to x (parallel parts): C(x) = (ε₀ w)/(d)[l + (K-1)x]. Charge constant: U = Q²/2C, so F = -dU/dx = (Q²)/(2C²)dC/dx = (Q²)/(2C²)·(ε₀ w(K-1))/(d). Near the edge x ≈ 0: C = (ε₀ w l)/(d), so F = (Q² d)/(2wl²ε₀)(K-1). So the answer is (c).

A parallel plate capacitor is made of two plates of length l, width w… | JEE Main 2014 (11 Apr, Shift 1)

Q108JEE Main 2014 · 11 APR · SHIFT 1Difficult

Dielectrics

A parallel plate capacitor is made of two plates of length l, width w and separated by distance d. A dielectric slab (dielectric constant K) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force F = $-\frac{\partial U}{\partial x}$ where U is the energy of the capacitor when dielectric is inside the capacitor up to distance x (see figure). If the charge on the capacitor is Q then the force on the dielectric when it is near the edge is:

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Answer: (c)

With slab in up to $x$ (parallel parts): $C(x) = \frac{\epsilon_0 w}{d}[l + (K-1)x]$ .
Charge constant: $U = \frac{Q^2}{2C}$ , so $F = -\frac{dU}{dx} = \frac{Q^2}{2C^2}\frac{dC}{dx} = \frac{Q^2}{2C^2}\cdot\frac{\epsilon_0 w(K-1)}{d}$ .
Near the edge $x \approx 0$ : $C = \frac{\epsilon_0 w l}{d}$ , so $F = \frac{Q^2 d}{2wl^2\epsilon_0}(K-1)$ .
So the answer is (c).

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