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  • Can someone help me with this, - Electrostatics - BITSAT

 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 :

  • Option 1)

    \frac{Q^{2}d}{2wl^{2}\epsilon _{0}}K

  • Option 2)

    \frac{Q^{2}w}{2dl^{2}\epsilon _{0}}(K-1)

  • Option 3)

    \frac{Q^{2}d}{2wl^{2}\epsilon _{0}}(K-1)

  • Option 4)

    \frac{Q^{2}w}{2dl^{2}\epsilon _{0}}K

 

Answers (2)

best_answer

As we discussed in concept

If K filled between the plates -

{C}'=K\frac{\epsilon _{0}A}{d}={C}'=Ck

 

 

- wherein

C\propto A

C\propto\frac{1}{d}

 

 

 

 

 C= C_{1}+ C_{2}= \frac{K(x\omega )\varepsilon _{0}}{d} + \frac{(l-x)\omega \varepsilon _{0}}{d}

C=\frac{\omega \varepsilon _{0}}{d}\times (Kx + (l-x))

v=\frac{1}{2}\times \frac{Q^{2}}{C}= \frac{Q^{2}d}{2\omega \varepsilon _{0}(\varepsilon +(k-1)x)}

\frac{\partial v }{\partial x}=-\frac{dQ^{2}(K-1)}{2\omega \varepsilon _{0}(l+(k-1)x)^{2}}

F=-\frac{\partial v}{\partial x}= \frac{Q^{2}d(K-1)}{2\omega l^{2}\varepsilon _{0}} at x=0

 


Option 1)

\frac{Q^{2}d}{2wl^{2}\epsilon _{0}}K

Option 2)

\frac{Q^{2}w}{2dl^{2}\epsilon _{0}}(K-1)

Option 3)

\frac{Q^{2}d}{2wl^{2}\epsilon _{0}}(K-1)

Option 4)

\frac{Q^{2}w}{2dl^{2}\epsilon _{0}}K

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