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Question Number 34501 by rahul 19 last updated on 07/May/18

Commented by rahul 19 last updated on 07/May/18

$a n s . i s \frac{ε_{0} K_{1} K_{2} a^{2} \ln \frac{K_{1}}{K_{2}}}{(K_{1} - K_{2}) d} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/May/18
	$As per picture the length of metal plate is a and assume widgh is b,separation between is d.Now horizontally at a distance x we take a small strip dx where a≥x≥0 area of strip is b×dx let in this small plrtion of metal plates thicknesd of dielectric is y and d−y where d≥y≥0 so two capacitor are in series connection dC_1 =((ε_0 k_1 .bdx)/y)and dC_2 =((ε_0 k_2 bdx)/((d−y) )) tanθ=(d/a)=(y/x) net capacitance=dC (1/dC)=(1/(dC_1 ))+(1/dC_2 ) =(y/(ε_0 k_1 bdx))+((d−y)/(ε_0 k_2 bdx)) y=(d/a)x dy=(d/a)dx (1/dC)=((k_2 y+k_1 (d−y))/(ε_0 bk_1 k_2 dx)) dC=((ε_0 bk_1 k_2 dx)/(k_2 (d/a)x+k_1 (d−(d/a)x))) putting the value of y =((ε_0 bk_1 k_2 dx)/(k_1 d+(k_2 −k_1 )(d/a)x)) now intregate both side and put the limit ofx a≥x≥0 =((ε_0 bk_1 k_2 )/((k_2 −k_1 )(d/a)))∫_0 ^a (dx/(((k_1 d)/((k_2 −k_(1)) (d/a)))+x)) =((e_0 bk_1 k_2 )/((k_2 −k_1 )(d/a)))∣log_e {((k_1 d)/(k_2 −k_1 )(d/a)))+x}∣_0 ^a =((e_0 bk_1 k_2 )/((k_2 −k_1 )(d/a)))∣log_e {((k_1 d)/((k_2 −k_1 )(d/a)))+a}−log((k_1 a)/(k_2 −k_1 ))∣ =((e_0 bk_1 k_2 a)/((k_2 −k_1 )d))∣log{((k_1 d+k_2 d−k_1 d)/((k_2 −k_1 )(d/a)))}−log((k_1 a)/(k_2 −k_1 ))∣ =((e_0 abk_1 k_2 )/((k_2 −k_1 )d))∣log((k_2 a)/((k_2 −k_1 )))−log((k_1 a)/((k_2 −k_1 )))∣ =((e_0 abk_1 k_2 )/((k_2 −k_1 )d)).log(k_2 /k_1 )$
	$A s p e r p i c t u r e t h e l e n g t h o f m e t a l p l a t e i s a$ $a n d a s s u m e w i d g h i s b, s e p a r a t i o n b e t w e e n i s$ $d . N o w h o r i z o n t a l l y a t a d i s t a n c e x w e t a k e a$ $s m a l l s t r i p d x w h e r e a ⩾ x ⩾ 0$ $a r e a o f s t r i p i s b \times d x$ $l e t i n t h i s s m a l l p l r t i o n o f m e t a l p l a t e s$ $t h i c k n e s d o f d i e l e c t r i c i s y a n d d - y$ $w h e r e d ⩾ y ⩾ 0$ $s o t w o c a p a c i t o r a r e i n s e r i e s c o n n e c t i o n$ ${d C}_{1} = \frac{ϵ_{0} k_{1} . b d x}{y} a n d {d C}_{2} = \frac{ϵ_{0} k_{2} b d x}{(d - y)}$ $t a n θ = \frac{d}{a} = \frac{y}{x} n e t c a p a c i t a n c e = d C$ $\frac{1}{d C} = \frac{1}{{d C}_{1}} + \frac{1}{{d C}_{2}}$ $= \frac{y}{ϵ_{0} k_{1} b d x} + \frac{d - y}{ϵ_{0} k_{2} b d x}$ $y = \frac{d}{a} x$ $d y = \frac{d}{a} d x$ $\frac{1}{d C} = \frac{k_{2} y + k_{1} (d - y)}{ϵ_{0} {b k}_{1} k_{2} d x}$ $d C = \frac{ϵ_{0} {b k}_{1} k_{2} d x}{k_{2} \frac{d}{a} x + k_{1} (d - \frac{d}{a} x)} p u t t i n g t h e v a l u e o f y$ $= \frac{ϵ_{0} {b k}_{1} k_{2} d x}{k_{1} d + (k_{2} - k_{1}) \frac{d}{a} x}$ $n o w i n t r e g a t e b o t h s i d e a n d p u t t h e l i m i t o f x$ $a ⩾ x ⩾ 0$ $= \frac{ϵ_{0} {b k}_{1} k_{2}}{(k_{2} - k_{1}) \frac{d}{a}} \underset{0}{\int^{a}} \frac{d x}{\frac{k_{1} d}{(k_{2} - k_{1)} \frac{d}{a}} + x}$ $= \frac{e_{0} {b k}_{1} k_{2}}{(k_{2} - k_{1}) \frac{d}{a}} ∣ {l o g}_{e} {\frac{k_{1} d}{k_{2} - k_{1}) \frac{d}{a}} + x} \underset{0}{\overset{a}{∣}}$ $= \frac{e_{0} {b k}_{1} k_{2}}{(k_{2} - k_{1}) \frac{d}{a}} ∣ {l o g}_{e} {\frac{k_{1} d}{(k_{2} - k_{1}) \frac{d}{a}} + a} - l o g \frac{k_{1} a}{k_{2} - k_{1}} ∣$ $= \frac{e_{0} {b k}_{1} k_{2} a}{(k_{2} - k_{1}) d} ∣ l o g {\frac{k_{1} d + k_{2} d - k_{1} d}{(k_{2} - k_{1}) \frac{d}{a}}} - l o g \frac{k_{1} a}{k_{2} - k_{1}} ∣$ $= \frac{e_{0} {a b k}_{1} k_{2}}{(k_{2} - k_{1}) d} ∣ l o g \frac{k_{2} a}{(k_{2} - k_{1})} - l o g \frac{k_{1} a}{(k_{2} - k_{1})} ∣$ $= \frac{e_{0} {a b k}_{1} k_{2}}{(k_{2} - k_{1}) d} . l o g \frac{k_{2}}{k_{1}}$
	Commented by rahul 19 last updated on 07/May/18

	$T h a n k y o u s i r .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 08/May/18

	$g i v e n p r o b l e m p l a t e i s s q u a r e s o a = b$ $s o r e p l a c e b b y a$
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