Derivations of 3 chapters

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IIT JEE Mains/Advanced, NEET, BITSAT, CBSE Board Himanshu Bhandari (B.Tech, IIT G) +91 9867530456
Electric field at a point on the axial position of an electric dipole

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Electric field at a point on the equatorial position of an electric dipole

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Energy stored in a capacitor and Energy stored per unit volume

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Effective capacitance when capacitors are connected in series

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Effective capacitance when capacitors are connected in parallel

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Capacitance of Parallel Plate Capacitor

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Electric Field due to Linear Charge

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Electric Field due to Plane Sheet
For an infinite sheet of charge, the electric field will be perpendicular to
the surface. Therefore only the ends of a cylindrical Gaussian surface will
contribute to the electric flux . In this case a cylindrical Gaussian surface
perpendicular to the charge sheet is used. The resulting field is half that of a
conductor at equilibrium with this surface charge density.

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Electric field between two parallel plates

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Electric field due to thin Spherical Shell

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Electric field inside a Charged Sphere

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Torque on an electric dipole in a uniform electric field

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Potential Energy of Dipole

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Electric Field and Potential Relation

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Potential Energy between two point charges

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Potential due to multiple charges

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Capacitor with a Dielectric Slab

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Capacitor with a Metallic Sheet

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Drift Velocity Derivation

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Relation between Drift velocity and Current

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Relation between Current Density and Electric Field

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Derivation of Cells in Series
Consider the points A, B and C and let V (A), V (B) and V (C) be the potentials of these points respectively.
V (A) - V (B) will be the potential difference between the positive and negative terminals for the first cell.
So VAB = V (A) - V (B) = ε1- Ir1.
VBC = V (B) - V (C) = ε2 – Ir2.
Now the potential difference between the terminals A and C is
VAC = V (A) – V(C) = [V (A) - V (B)] + V (B) - V (C)]
= ε1- Ir1 + ε2 – Ir2
= ( ε1 + ε2) – I(r1+r2).
In case if we replace this combination of cells by a single cell between the points A and C with emf εeq and internal
resistance req, VAC = εeq - req. and thus we found out that εeq = ε1 + ε2 and req = r1+r2

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Derivation of Cells in Parallel
V = V (B1) - V (B2) = ε1- I1 r1. The point B1 and B2 are connected similar to the second cell.
V = V (B1) - V (B2) = ε2 – I2 r2. By ohm’s law we know that I = V / R. Now substitute these values in the
equation
If we replace the cells by a single cell lying
between the point B1 and B2 with emf εeq and internal resistance req, then
V = εeq - Ireq.
It is the same as when we connect the resistors in parallel connection. For n number of cells
connected in parallel with emf ε1, ε2…… εn and internal resistance r1, r2…. rn

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Heat Produced in a Resistor

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Wheatstone Bridge and Meter Bridge
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Derivations of 3 chapters

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Derivations of 3 chapters