JEE Main 2014 · 11 Apr · Shift 1 · Physics Q17

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PhysicsSingle correct+4 / -1Question 17 of 90

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 $F=-\dfrac{\partial U}{\partial x}$ 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 :

Two vertical plates of length l, separation d; a hatched dielectric slab is above them, entering a distance x between the plates (dotted).

\dfrac{Q^2d}{2wl^2\epsilon_o}K

\dfrac{Q^2w}{2dl^2\epsilon_o}(K-1)

\dfrac{Q^2d}{2wl^2\epsilon_o}(K-1)

\dfrac{Q^2w}{2dl^2\epsilon_o}K

Answer and worked solution

Official answer: C

With $Q$ fixed, $U=\dfrac{Q^2}{2C}$ and $C(x)=\dfrac{\epsilon_ow}{d}\left[(l-x)+Kx\right]$ . So
$F=\frac{Q^2}{2C^2}\frac{dC}{dx}=\frac{Q^2}{2C^2}\cdot\frac{\epsilon_ow(K-1)}d.$
Near the edge, $x\approx0$ and $C=\dfrac{\epsilon_owl}d$ :
$F=\frac{Q^2d}{2wl^2\epsilon_o}(K-1).$

Topics:Dielectrics Energy stored in a capacitor

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 17