physicscatalyst.com logo




Capacitance






3. Parallel plate capacitor



  • A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance shown below in the figure 3.
    Parallel plate capacitor

  • Suppose two plates of the capacitor has equal and opposite charge Q on them. If A is the area of each plate then surface charge density on each plate is
              σ=Q/A
  • We have already calculated field between two oppositely charged plates using gauss's law which is
              E=σ/ε0=Q/ε0A
    and in this result effects near the edges of the plates have been neglected.
  • Since electric field between the plates is uniform the potential difference between the plates is
              V=Ed=Qd/ε0A
    where , d is the separation between the plates.
  • Thus, capacitance of parallel plate capacitor in vacuum is
              C=Q/V=ε0A/d               (3)
  • From equation 3 we see that quantities on which capacitance of parallel plate capacitor depends i.e.,ε0 , A and d are all constants for a capacitor.
  • Thus we see that in this case capacitance is independent of charge on the capacitor but depends on area of it's plates and separation distance between the plates.

4.Cylinderical capacitor



  • A cylinderical capacitor is made up of a conducting cylinder or wire of radius a surrounded by another concentric cylinderical shel of radius b (b>a).

  • Let L be the length of both the cylinders and charge on inner cylender is +Q and charge on outer cylinder is -Q.
  • For calculate electric field between the conductors using Gauss's law consider a gaussian consider a gaussian surface of radius r and length L1 as shown in figure 4.

    Cylinderical capacitor

  • According to Gauss's law flux through this surface is q/ε0 where q is net charge inside this surface.
  • We know that electric flux is given by
              φ=E.A
                =EAcosθ     
                =EA

    since electric field is constant in magnitude on the gaussian surface and is perpandicular to this surface. Thus,
               φ=E(2πrL)
    since            φ=q/ε0
    =>           E(2πrL)=(λL)/ε0

    where λ = Q/L = charge per unit length
    =>           $$E=\frac{\lambda }{2\pi \epsilon _{0}r}$$
                                  (4)
  • If potential at inner cylinder is Va and Vb is potential of outer cylinder then potential difference between both the cylinders is
              V=Va and           Vb=∫Edr
    where limits of integration goes from a to b.


  • Potential of inner conductor is greater then that of outer conductor because inner cylinder carries positive charge. Thus potential difference is $$V=\frac{Qln(b/a)}{2\pi \epsilon _{0}L}$$
  • Thus capacitance of cylinderical capacitor is
              C=Q/V
    or, $$C=\frac{2\pi \epsilon _{0}L{(b/a))}$$
                                       (5)
  • From equation 5 it can easily be concluded that capacitance of a cylinderical capacitor depends on length of cylinders.
  • More is the length of cylinders , more charge could be stored on the capacitor for a given potential difference.



5. Spherical capacitor



  • A spherical capacitor consists of a solid or hollow spherical conductor of radius a , surrounded by another hollow concentric spherical of radius b shown below in figure 5
    Spherical capacitor

  • Let +Q be the charge given to the inner sphere and -Q be the charge given to the outer sphere.
  • The field at any point between conductors is same as that of point charge Q at the origin and charge on outer shell does not contribute to the field inside it.
  • Thus electric field between conductors is $$E=\frac{Q}{2\pi \epsilon _{0}r^{2}}$$
              
  • Potential difference between two conductors is
         V=Va-Vb
         =-∫E.dr

    where limits of integration goes from a to b.
    On integrating we get potential difference between to conductors as
    $$V=\frac{Q(b-a)}{4\pi \epsilon _{0}ba}$$
  • Now , capacitance of spherical conductor is
         C=Q/V
    or, $$C=\frac{4\pi \epsilon _{0}ba}{(b-a)}$$
                                  (6)
  • again if radius of outer conductor aproaches to infinity then from equation 6 we have
         C=4πε0a          (7)
  • Equation 7 gives the capacitance of single isolated sphere of radius a.
  • Thus capacitance of isolated spherical conductor is proportional to its radius.




link to this page by copying the following text


Class 12 Maths Class 12 Physics





Note to our visitors :-

Thanks for visiting our website. From feedback of our visitors we came to know that sometimes you are not able to see the answers given under "Answers" tab below questions. This might happen sometimes as we use javascript there. So you can view answers where they are available by reloding the page and letting it reload properly by waiting few more seconds before clicking the button.
We really do hope that this resolve the issue. If you still hare facing problems then feel free to contact us using feedback button or contact us directly by sending is an email at [email protected]
We are aware that our users want answers to all the questions in the website. Since ours is more or less a one man army we are working towards providing answers to questions available at our website.