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Question-34501




Question Number 34501 by rahul 19 last updated on 07/May/18
Commented by rahul 19 last updated on 07/May/18
ans. is    ((ε_0 K_1 K_2 a^2 ln (K_1 /K_2 ))/((K_1 −K_2 )d)) .
$${ans}.\:{is}\:\:\:\:\frac{\varepsilon_{\mathrm{0}} {K}_{\mathrm{1}} {K}_{\mathrm{2}} {a}^{\mathrm{2}} \mathrm{ln}\:\frac{{K}_{\mathrm{1}} }{{K}_{\mathrm{2}} }}{\left({K}_{\mathrm{1}} −{K}_{\mathrm{2}} \right){d}}\:. \\ $$
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 )
$${As}\:{per}\:{picture}\:{the}\:{length}\:{of}\:{metal}\:{plate}\:{is}\:{a}\: \\ $$$${and}\:{assume}\:{widgh}\:{is}\:{b},{separation}\:{between}\:{is} \\ $$$${d}.{Now}\:{horizontally}\:{at}\:{a}\:{distance}\:{x}\:{we}\:{take}\:{a}\: \\ $$$${small}\:{strip}\:{dx}\:{where}\:\:{a}\geqslant{x}\geqslant\mathrm{0} \\ $$$$\:{area}\:{of}\:{strip}\:{is}\:{b}×{dx} \\ $$$${let}\:{in}\:{this}\:{small}\:{plrtion}\:{of}\:{metal}\:{plates}\: \\ $$$${thicknesd}\:{of}\:{dielectric}\:{is}\:\:{y}\:{and}\:{d}−{y} \\ $$$${where}\:{d}\geqslant{y}\geqslant\mathrm{0} \\ $$$${so}\:{two}\:{capacitor}\:{are}\:{in}\:{series}\:{connection} \\ $$$${dC}_{\mathrm{1}} =\frac{\epsilon_{\mathrm{0}} {k}_{\mathrm{1}} .{bdx}}{{y}}{and}\:{dC}_{\mathrm{2}} =\frac{\epsilon_{\mathrm{0}} {k}_{\mathrm{2}} {bdx}}{\left({d}−{y}\right)\:} \\ $$$${tan}\theta=\frac{{d}}{{a}}=\frac{{y}}{{x}}\:{net}\:{capacitance}={dC} \\ $$$$\frac{\mathrm{1}}{{dC}}=\frac{\mathrm{1}}{{dC}_{\mathrm{1}} \:}+\frac{\mathrm{1}}{{dC}_{\mathrm{2}} } \\ $$$$=\frac{{y}}{\epsilon_{\mathrm{0}} {k}_{\mathrm{1}} {bdx}}+\frac{{d}−{y}}{\epsilon_{\mathrm{0}} {k}_{\mathrm{2}} {bdx}} \\ $$$${y}=\frac{{d}}{{a}}{x} \\ $$$${dy}=\frac{{d}}{{a}}{dx} \\ $$$$\frac{\mathrm{1}}{{dC}}=\frac{{k}_{\mathrm{2}} {y}+{k}_{\mathrm{1}} \left({d}−{y}\right)}{\epsilon_{\mathrm{0}} {bk}_{\mathrm{1}} {k}_{\mathrm{2}} {dx}} \\ $$$${dC}=\frac{\epsilon_{\mathrm{0}} {bk}_{\mathrm{1}} {k}_{\mathrm{2}} {dx}}{{k}_{\mathrm{2}} \frac{{d}}{{a}}{x}+{k}_{\mathrm{1}} \left({d}−\frac{{d}}{{a}}{x}\right)}\:{putting}\:{the}\:{value}\:{of}\:{y} \\ $$$$=\frac{\epsilon_{\mathrm{0}} {bk}_{\mathrm{1}} {k}_{\mathrm{2}} {dx}}{{k}_{\mathrm{1}} {d}+\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)\frac{{d}}{{a}}{x}} \\ $$$${now}\:{intregate}\:{both}\:{side}\:{and}\:{put}\:{the}\:{limit}\:{ofx} \\ $$$$\:{a}\geqslant{x}\geqslant\mathrm{0} \\ $$$$=\frac{\epsilon_{\mathrm{0}} {bk}_{\mathrm{1}} {k}_{\mathrm{2}} }{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)\frac{{d}}{{a}}}\underset{\mathrm{0}} {\overset{{a}} {\int}}\frac{{dx}}{\frac{{k}_{\mathrm{1}} {d}}{\left({k}_{\mathrm{2}} −{k}_{\left.\mathrm{1}\right)} \frac{{d}}{{a}}\right.}+{x}} \\ $$$$=\frac{{e}_{\mathrm{0}} {bk}_{\mathrm{1}} {k}_{\mathrm{2}} }{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)\frac{{d}}{{a}}}\mid{log}_{{e}} \left\{\frac{{k}_{\mathrm{1}} {d}}{\left.{k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)\frac{{d}}{{a}}}+{x}\right\}\underset{\mathrm{0}} {\overset{{a}} {\mid}} \\ $$$$=\frac{{e}_{\mathrm{0}} {bk}_{\mathrm{1}} {k}_{\mathrm{2}} }{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)\frac{{d}}{{a}}}\mid{log}_{{e}} \left\{\frac{{k}_{\mathrm{1}} {d}}{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)\frac{{d}}{{a}}}+{a}\right\}−{log}\frac{{k}_{\mathrm{1}} {a}}{{k}_{\mathrm{2}} −{k}_{\mathrm{1}} }\mid \\ $$$$=\frac{{e}_{\mathrm{0}} {bk}_{\mathrm{1}} {k}_{\mathrm{2}} {a}}{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right){d}}\mid{log}\left\{\frac{{k}_{\mathrm{1}} {d}+{k}_{\mathrm{2}} {d}−{k}_{\mathrm{1}} {d}}{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)\frac{{d}}{{a}}}\right\}−{log}\frac{{k}_{\mathrm{1}} {a}}{{k}_{\mathrm{2}} −{k}_{\mathrm{1}} }\mid \\ $$$$=\frac{{e}_{\mathrm{0}} {abk}_{\mathrm{1}} {k}_{\mathrm{2}} }{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right){d}}\mid{log}\frac{{k}_{\mathrm{2}} {a}}{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)}−{log}\frac{{k}_{\mathrm{1}} {a}}{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right)}\mid \\ $$$$=\frac{{e}_{\mathrm{0}} {abk}_{\mathrm{1}} {k}_{\mathrm{2}} }{\left({k}_{\mathrm{2}} −{k}_{\mathrm{1}} \right){d}}.{log}\frac{{k}_{\mathrm{2}} }{{k}_{\mathrm{1}} } \\ $$$$ \\ $$
Commented by rahul 19 last updated on 07/May/18
Thank you sir.
$${Thank}\:{you}\:{sir}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 08/May/18
given problem plate is square so a=b    so replace b by a
$${given}\:{problem}\:{plate}\:{is}\:{square}\:{so}\:{a}={b}\:\: \\ $$$${so}\:{replace}\:{b}\:{by}\:{a} \\ $$

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