Ls 2 - Electrostatic Potential and Capacitance
Notes
Questions
A regular hexagon of side 10cm has a charge 5uC at each of its vertices . Calculate the potential at the centre of the hexagon.
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
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

Electric potential: V=qW
W = q(VA−VB) where VA>VB
E = −dldV (relation between electric intensity and potential)
Electric potential at point due to isolated point charge, V=4πεo1rq
Electric potential at point due to an electric dipole, V=4πεo1r2p cosθ
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
Potential energy of system of charges in the presence of external electric field
Single charge: - U = Vq
Two charges: -U=V(r1)q1+V(r2)q2+4πεo1r12q1q2
Electric dipole: U=−pE(cosθo−cosθ1)
C=VQ
Capacitance of parallel plate capacitor: V=dεrεoA
Energy stored in capacitor, E=21CV2
Series combination of capacitors, -= +1+- C1C2
For two capacitors connected in series, Cs = +C2
For n identical capacitors connected in series, Cn =-
Parallel combination of capacitors, Cp = C1 + C2+ C3
For two Capacitors connected in parallel, Cp = C1+ C2
For n identical capacitors connected in parallel, Cp = nC
Dielectric polarization P = x E where x is electric susceptibility
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