Question and Answers Forum

All Questions      Topic List

Electrostatics Questions

Previous in All Question      Next in All Question      

Previous in Electrostatics      Next in Electrostatics      

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

$${given}\:{problem}\:{plate}\:{is}\:{square}\:{so}\:{a}={b}\:\: \\ $$$${so}\:{replace}\:{b}\:{by}\:{a} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com