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Additivity of charges:
- Point charges are scalars and can be added algebraically. If q1, q2, q3, … qn, are point charges, the total charge qtot=q1+ q2 + q3+ qn
- Charges have no direction but can be positive or negative.
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Raoult’s law
- Raoult’s law establishes a quantitative relationship between the partial vapour pressure and mole fraction of a solution.
- This law is only for liquid-liquid solution.
- The law states that for a solution of volatile liquids, the partial vapour pressure (p) of each component in the solution is directly proportional to its mole fraction (x).
- Mathematically, p ∝ x
Or p = p 0x
Where p0 is the vapour pressure of pure component at the same temperature.
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Electric potential energy is defined as the total potential energy a unit charge will possess if located at any point in the outer space.
If two charges q1 and q2 are separated by a distance d, the electric potential energy of the system is;
U = 1/(4πεo) × [q1q2/d]
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Gauss’s Law
According to Gauss’s law, the total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The total electric flux through a closed surface is zero if no charge is enclosed by the surface.
- Gauss’s law is true for any closed surface, no matter what its shape or size.
- The term q on the right side of Gauss’s lawincludes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.
- In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field [whose flux appears on the left side of Eq. (1.31)] is due to all the charges, both inside and outside S. The term q on the right side of Gauss’s law, however, represents only the total charge inside S.
- The surface that we choose for the application of Gauss’s law is called the Gaussian surface. The Gaussian surface can pass through a continuous charge distribution.
- Gauss’s law is useful for the calculation of the electrostatic field for a symmetric system.
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Yogita Ingle 5 years, 5 months ago
Given Power Factor = cos (phi) = 0.5
So (phi) = 60 degrees
The phase difference between the voltage and the current = phi = 60 degrees Ans
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1. In the case of a charged ring of radius R on its axis at a distance x from the centre of the ring. E = <latex>4π∈01(R2+x2)3/2qx</latex>. At the centre, x = 0 and E = 0.
2. In case of an infinite line of charge, at a distance ‘r’. E = (1/4 × πrε0) (2π/r) = λ/2πrε0. Where λ is the linear charge density.
3. The intensity of the electric field near a plane sheet of charge is E = σ/2ε0K where σ = surface charge density.
4. The intensity of the electric field near a plane charged conductor E = σ/Kε0 in a medium of dielectric constant K. If the dielectric medium is air, then Eair = σ/ε0.
5. The field between two parallel plates of a condenser is E = σ/ε0, where σ is the surface charge density.
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