IIT JEE , AIEEE Forum - Physics , Electricity - h.c verma ch capacitors pg no.170 ques 68,67

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h.c verma ch capacitors pg no.170 ques 68,67

iitkgp_bipin

Please post the complete question and that too one at a time.

indrajeet_bariar

SOLUTION TO Q.67

see to solve this question u must know the expression for force experienced by a dielectric slab kept inside a capacitor. Suppose a capacitor with length and width of the plated l and b respectively. the seperation between the plated is d. now a dielectric slab of dielectric constant K is kept between the plated of the capacitor such tht x length is inside the capacitor and the width of the dielectric is also d. then the dielectric is pulled inside the capacitor with a force given by

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

now let the length of the dielectric be 2x so tht x length remains inside each capacitor. now for the slab to be in equilibrium the forces exerted by both the capacitors must be equal in magnitude. applying the above formula

$\frac{\epsilon_{0} b{E_{1}}^{2} (K_{1} - 1)}{2x} = \frac{\epsilon_{0} b{E_{2}}^{2} (K_{2} - 1)}{2x}$

solving this u will get

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

i will soon post solution to q.68 coz. its very tedious to type the complete solution here.

PLZ... RATE ME IF U UNDERSTOOD THE SOLUTION CLEARLY OR IF IN DOUBT U CAN ASK FURTHER.

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