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Spherical capacitor





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
    $V=V_a -V_b$
    $=- \int 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= \frac {Q}{V} $
    or,
    $C=\frac{4\pi \epsilon _{0}ba}{(b-a)}$ ----(1)
  • again if radius of outer conductor approaches to infinity then from equation 6 we have
    $C=4 \pi \epsilon _{0} a$ ----(2)
  • Equation 2 gives the capacitance of single isolated sphere of radius a.
  • Thus capacitance of isolated spherical conductor is proportional to its radius.

Spherical capacitor when inner sphere is earthed

  • If a positive charge of Q coulombs is given to the outer sphere B, it will distribute itself over both its inner and outer surfaces.
  • Let the charges of $Q_1$ and $Q_2$ coulombs be at the inner and outer surfaces respectively of sphere B where $Q = Q_1 +Q_2$,
  • The charge + $Q_1$ on the inner surface of outer sphere B will induce a charge of -$Q_1$ coulombs on the outer surface of inner sphere A and + $Q_1$ coulombs on the inner surface of sphere A, which will go to earth.
  • Now there are two capacitors connected in parallel.
    (i) One capacitor consists outer surface of sphere B and earth having capacitance $C_1 = 4 \pi \epsilon _0 b$ farads
    (ii) Second capacitor consisting of inner surface of outer sphere B and the outer surface of inner sphere A having capacitance
    $C_2=\frac{4\pi \epsilon _{0}ba}{(b-a)}$
  • Final Capacitance
    $C=C_1+C_2=4 \pi \epsilon _0 b + \frac{4\pi \epsilon _{0}ba}{(b-a)}= \frac{4\pi \epsilon _{0}b^2}{(b-a)}$
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Class 12 Maths Class 12 Physics





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