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Find the equivalent capacitance of the given figure. |
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Answer» Solution :`1/(DC)=(y)/(epsi_(0)k_(1)BDX)+(d_(0)-y)/(epsi_(0)k_(2)bdx)` `1/(dc)=(d_(0)k_(1)+y(k_(2)-k_(1)))/(epsi_(0)k_(1)k_(2)bdx)` `dx=(epsi_(0)k_(1)k_(2)bdx)/(d_(0)k_(1)+y(k_(2)-k_(1)))` All these capacitors (SMALL) are parallel So, `C_(eq)=underset(0)OVERSET(C_(eq))int dC=underset(0)overset(a)int (epsi_(0)k_(1)k_(2)bdx)/(d_(0)k_(1)+y(k_(2)-k_(1)))` Now, `d_(0)/a=y/x rArr C_(eq)=underset(0)overset(a)int (epsi_(0)k_(1)k_(2)bdx)/(d_(0)k_(1)+d_(0)/AX(k_(2)-k_(1)))` `C_(eq)=(a epsi_(0)k_(1)k_(2)b)/(d_(0)(k_(2)-k_(1)))ln [((k_(2)-k_(1)))/(k_(1))]` 
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