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  • A parallel plate capacitor of capacitance C is charged to a potential V by a battery. Without disconnecting the battery, the distance between the plates is tripled and a dielectric medium of k = 10 is introduced between the plates of the capacitor. Exp

A parallel plate capacitor of capacitance C is charged to a potential V by a battery. Without disconnecting the battery, the distance between the plates is tripled and a dielectric medium of k = 10 is introduced between the plates of the capacitor. Explain giving reasons, how will the following be affected :
(i) capacitance of the capacitor
(ii) charge on the capacitor, and
(iii) energy density of the capacitor.

 

 

 
 
 
 
 

Answers (1)

(i) Capacitance
Case 1
            C= \frac{\varepsilon _{0}A}{d}
case 2
        {C}'=\frac{k \varepsilon _{0}A}{{d}'}
             = \frac{10\, \varepsilon _{0}A}{3d}
            = \frac{10}{3}C
Case 1
ii)
  charge Q= CV
case 2
charge {Q}'= {C}'V= \frac{10}{3}Q
iii) The energy density of the capacitor
Case 1     V= \frac{1}{2}\, \varepsilon _{0}\, E^{2}
case 2 {V}'= \frac{1}{2}k\, \varepsilon _{0}\, {E}'\, ^{2}
         = \frac{1}{2}\times 10\, \varepsilon _{0}\left ( \frac{V^{2}}{l^{2}} \right )
        = \frac{1}{2}\times 10\,\times \varepsilon _{0}\left ( \frac{V^{2}}{9d^{2}} \right )
      = \frac{10}{9}\left ( \frac{1}{2}\varepsilon _{0} \frac{V^{2}}{d^{2}}\right )
     = \frac{10}{9}\; \frac{1}{2}\, \varepsilon _{0}\: E^{2}
{V}'=\frac{10}{9}\, V

Posted by

Safeer PP

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