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  1. class
  2. Class 12
  3. Physics
  4. Lessons

Ls 2 - Electrostatic Potential and Capacitance

Last updated 10 months ago

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Notes

Questions

  1. A regular hexagon of side 10cm has a charge 5uC at each of its vertices . Calculate the potential at the centre of the hexagon.

  2. As in the figure, if a capacitor of capacitance 'C' is charged by connecting it with resistance 'R', then energy given by the battery will be a) 1/2CV^2 b) less than 1/2CV^2 c) CV^2 d) more than CV^2

  3. There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between in the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60 deg with the direction of the field? (A) 589.5 V (B) 589.2 V (C) 589.4 V (D) 589.6 V

Worksheet

Notes

Formula Cheat sheet

  1. Potential energy of system of charges in the absence of external electric field - is due to superposition principle which is very similar to addition of vectors

  2. Potential energy of system of charges in the presence of external electric field

    1. Single charge: - U = Vq

  3. Series combination of capacitors, -= +1+- C1C2

    1. For two capacitors connected in series, Cs = +C2

    2. For n identical capacitors connected in series, Cn =-

  4. Dielectric polarization P = x E where x is electric susceptibility

Electric potential: V=WqV = \frac{W}{q}V=qW​

W = q(VAβˆ’VB)Β whereΒ VA>VBq (V_A-V_B) \ where \ V_A > V_Bq(VAβ€‹βˆ’VB​)Β whereΒ VA​>VB​

E = βˆ’dVdl- \frac{dV}{dl}βˆ’dldV​ (relation between electric intensity and potential)

Electric potential at point due to isolated point charge, V=14πΡoqrV = \frac{1}{4πΡ_o} \frac{q}{r}V=4πΡo​1​rq​

Electric potential at point due to an electric dipole, V=14πΡopΒ cosΞΈr2V = \frac{1}{4πΡ_o} \frac{p \ cos \theta}{r^2}V=4πΡo​1​r2pΒ cosθ​

Two charges: -U=V(r1)q1+V(r2)q2+14πΡoq1q2r12U= V(r_1)q_1+ V(r_2)q_2+ \frac{1}{4πΡ_o} \frac{q_1q_2}{r_{12}}U=V(r1​)q1​+V(r2​)q2​+4πΡo​1​r12​q1​q2​​

Electric dipole: U=βˆ’pE(cosΞΈoβˆ’cosΞΈ1)U =- pE (cos ΞΈ_o-cos ΞΈ_1)U=βˆ’pE(cosΞΈoβ€‹βˆ’cosΞΈ1​)

C=QVC= \frac{Q}{V}C=VQ​

Capacitance of parallel plate capacitor: V=Ξ΅rΞ΅oAdV = \frac{Ξ΅_rΞ΅_oA}{d}V=dΞ΅r​Ρo​A​

Energy stored in capacitor, E=12CV2E = \frac{1}{2}CV^2E=21​CV2

Parallel combination of capacitors, CpC_pCp​ = C1 + C2+ C3

For two Capacitors connected in parallel, CpC_pCp​ = C1+ C2

For n identical capacitors connected in parallel, CpC_pCp​ = nC

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Electric Potential - Ls 2 - Physics -Rao Sir notes.pdf
Chapter 02 Book - Rao Sir - Worked Out.pdf
Electric Potential - PYQs.pdf
Ls 2 - Physics - Sugeeth Notes - 12 A - 01-07-2024.pdf - Google Drive