Electrostatic Potential and Capacitance

Capacitors and Capacitance, Parallel Plate Capacitor and Effect of Dielectric

  • The increase in potential is directly proportional to the change given to the conductor Q∝v.
  • The capacity of a conductor is numerically equal to the Electric charge which raises its potential by unit.
  • Capacity is independent of charge and potential of the conductor.
  • The relation between Q, C and V is Q = CV
  • Electric capacity of a spherical conductor is given by c = 4πε0R. {R = Radins}.
  • The capacity of a conductor depends on the surface area and the medium around the conductor (Dielectric constant of medium)
  • Energy stored in a capacitor \tt U=\frac{1}{2}\ CV^{2}
  • \tt U=\frac{Q^{2}}{2C}=\frac{1}{2}\ QV=\frac{1}{2}\ CV^{2}
  • When two charged bodies one connected by a conducting wire the charge flows from a conductor at higher potential to that at lower potential
  • When two spheres of capacities C1 & C2 with charges Q1 and Q2 one connected then new charges Q'1 : Q'2 = C1 : C2
  • Common potential when connected \tt v=\frac{Q_{1}+Q_{2}}{C_{1}+C_{2}}=\frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}
  • Effective capacitance \tt C=\frac{Q_{1}^{'}+Q_{2}^{'}}{v}=C_{1}+C_{2}
  • Loss in Energy of the system \tt \Delta U=\frac{1}{2}\ \frac{C_{1}C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^2
  • Capacitor is a device used to Electric charge and Electrostatic Energy at low potentials.
  • Capacity of a parallel plate capacitor in vacuum \tt C_{0}=\frac{\varepsilon_{0}A}{d}
  • Capacity of a parallel plate capacitor in medium \tt C=K\frac{\varepsilon_{0}A}{d}
  • \tt C=K\frac{\varepsilon_{0}A}{d}=\varepsilon_{r}\frac{\varepsilon_{0} A}{d}=\frac{\varepsilon A}{d}
  • Electric field between the plates is uniform \tt E=\frac{\sigma}{\varepsilon_{0}}
  • Potential difference between the plates \tt V=\frac{\sigma}{\varepsilon_{0}}\cdot d
  • Force on each plate of capacitor \tt F=\frac{1}{2}\ EQ
  • Capacitence when a dielectric slab of thickness ‘t’ is introduced between plates \tt C'=\frac{\varepsilon_{0}\ A}{d-t+\frac{t}{k}}=\frac{\varepsilon_{0}\ A}{d-t\left(1-\frac{1}{k}\right)}
  • When a metal slab of thickness ‘t’ is introduced \tt C'=\frac{\varepsilon_{0}\ A}{d-t}
  • When number of dielectrics and used \tt C=\frac{\varepsilon_{0}\ A}{d-t_{1}\left(1-\frac{1}{K_{1}}\right)-t_{2}\left(1-\frac{1}{K_{2}}\right)-t_{3}\left(1-\frac{1}{K_{3}}\right)...}
  • When number of dielectric one completely filled \tt C=\frac{\varepsilon_{0}\ A}{\frac{t_{1}}{K_{1}}+\frac{t_{2}}{K_{2}}+....\frac{t_{n}}{K_{n}}}

Capacitor & Capacitance View the Topic in this video From 43:21 To 56:23

Disclaimer: Compete.etutor.co may from time to time provide links to third party Internet sites under their respective fair use policy and it may from time to time provide materials from such third parties on this website. These third party sites and any third party materials are provided for viewers convenience and for non-commercial educational purpose only. Compete does not operate or control in any respect any information, products or services available on these third party sites. Compete.etutor.co makes no representations whatsoever concerning the content of these sites and the fact that compete.etutor.co has provided a link to such sites is NOT an endorsement, authorization, sponsorship, or affiliation by compete.etutor.co with respect to such sites, its services, the products displayed, its owners, or its providers.

1. When dielectric is partially filled between the plates of a parallel plate capacitor then it's capacitance increases but potential difference decreases. To maintain the capacitance and potential difference of capacitor as before separation between the plates has to be increases say by d'. In such case
K = \frac{t}{t - d}

2. Parallel plate capacitor: It consists of two similar flat conducting plates, arranged parallel to one another, separated by a distance. Its capacitance is given by C = \frac{\varepsilon_{0}A}{d} (when air is between the plates)