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Yogita Ingle 6 years, 4 months ago
Potential difference is the difference in electric potential energy between two points. It is the amount of work done in moving a unit positive charge from one point to another point in an electric field. This is measured in volt.
Posted by Navneet Kumar 6 years, 4 months ago
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Yogita Ingle 6 years, 4 months ago
Limitations of Ohm's Law:
Ohm's law is not a fundamental law of nature. There are a number of commonly used circuit elements which do not obey this law called Non – Ohmic conductors.
Non – Ohmic conductors have one or more of the following properties:
1. Potential difference (V) depends on current (I) non-linearly.
2. The relation between V and I is non-unique, that is, for the same current I, there is more than one value of voltage V.
3. The relationship between V and I depends on the sign of V f or the same absolute value of V.
Sia ? 6 years, 4 months ago
The resistance of a material changes when the temperature changes. So when the resistance changes, the ratio of the voltage across the resistor and the current passing through it will not be a constant. We will not get the same numbers always. Non – linear elements and unilateral networks do not support ohm's law.
Posted by Sameer Khan 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Electric Field of Point Charge. The electric field of a point charge Q can be obtained by a straightforward application of Gauss' law. Considering a Gaussian surface in the form of a sphere at radius r, the electric field has the same magnitude at every point of the sphere and is directed outward. Let us consider charge +q be uniformly distributed over a spherical shell of radius R. Let E is to be obtained at Plying outside of spherical shell.
{tex}\because {/tex} 'E' at any point is radially outward (if charge q is positive) and has same magnitude at all points which lie at the same distance r from centre of spherical shell such that r > R.
Therefore, Gaussian surface is concentric sphere of radius r such that r > R.
By Gauss' theorem
{tex}\oint \mathbf { E } \cdot d \mathbf { S } = \frac { q } { \varepsilon _ { 0 } } \Rightarrow \oint E d S \cos 0 ^ { \circ } = \frac { q } { \varepsilon _ { 0 } }{/tex}
[{tex}\because {/tex} E and dS are along the same direction]
{tex}E \cdot \oint d S = \frac { q } { \varepsilon _ { 0 } }{/tex} [{tex}\because {/tex} Magnitude of E is same at every point on Gaussian surface]
{tex}E \times 4 \pi r ^ { 2 } = \frac { q } { \varepsilon _ { 0 } }{/tex}
{tex}\Rightarrow \quad E = \frac { 1 } { 4 \pi \varepsilon _ { 0 } } \cdot \frac { q } { r ^ { 2 } }{/tex}
Now, graph
Variation of E with r for a spherical shell in which charge q is distributed over its surface which shows that electric field inside a spherical shell is zero.
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ii) Greater the conductivity of the electrolyte, lesser is the internal resistance of the cell. i.e. internal resistance depends on the nature of the electrolyte.
iii) The internal resistance of a cell is inversely proportional to the common area of the electrodes dipping in the electrolyte.
iv) The internal resistance of a cell depends on the nature of the electrodes.
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Sia ? 6 years, 4 months ago
To derive Coulomb’s Law from gauss law or to find the intensity of electric field due to a point charge +q at any point in space using Gauss’s law ,draw a Gaussian sphere of radius r at the centre of which charge +q is located (Try to make the figure yourself).
All the points on this surface are equivalent and according to the symmetric consideration the electric field E has the same magnitude at every point on the surface of the sphere and it is radially outward in direction.Therefore, for a area element dS around any point P on the Gaussian surface both E and dS are directed radially outward,that is ,the angle between E and dS is zero.Therefore,
The flux passing through the area element dS ,that is,
d φ =E.dS= EdS cos 00=EdS
Hence, the total flux through the entire Gaussian sphere is obtained as,
Φ=∫EdS
Or φ=E∫dS
But ∫dS is the total surface area of the sphere and is equal to 4πr2,that is,
Φ=E(4πr2) (1)
But according to <a title="Gauss’s law for electrostatics" href="https://winnerscience.com/electromagnetic-field-theory/gauss-law-electrostatics-derivation/">Gauss’s law for electrostatics</a>
Φ=q/ε0 (2)
Where q is the charge enclosed within the closed surface
By comparing equation (1) and (2) ,we get
E(4πr2)=q/ε0
Or E=q/4πε0r2 (3)
The equation (3) is the expression for the magnitude of the intensity of electric field E at a point,distant r from the point charge +q.
In vector form, E=1/4πε0 q/r2 =1/4πε0qr/r3
In a second point charge q0be placed at the point at which the magnitude of E is computed ,then the magnitude of the force acting on the second charge q0would be
F=q0E
By substituting value of E from equation (3),we get
F=qoq/4πε0r2 (4)
The equation (4) represents the Coulomb’s Law and it is derived from gauss law.
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