Gauss's Law By Sanjay Pandey

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2. Gauss’s Law by Sanjay Pandey

Gauss’s Law is physically equivalent to Coulomb’s Law. Gauss’s Law is an alternative description that is more convenient/easier to apply in situations of high symmetry. Gauss’s Law deals with a quantity known as electric flux. To use Gauss’s Law requires understanding how to calculate electric flux. Gauss’s Law permits drawing some sweeping, interesting, and technologically useful conclusions about charged conductors in external fields Recall: Electric Field Lines

Density of lines proportional to electric field strength/magnitude. Electric field direction tangent to lines. Lines leave/start at positive charges, enter/end at negative charges, number proportional to magnitude of charge. Lines can never cross: only one line at any point, field has a unique direction. Flux of an electric field Electric flux is a quantitative way to make use of the behavior of electric field lines in a manner consistent with Coulomb’s Law. Loosely speaking, electric flux is the amount of “flow” of field lines through imaginary surfaces.

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Another ‘picture’ of flux

Flux is the “amount of stuff” going through an area.

area.

𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = 𝑣𝑣⃗. 𝐴𝐴⃗

The flux depends on the relative orientation of the “stuff going through” the area and the plane of the

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𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = 𝑣𝑣⃗. 𝐴𝐴⃗⊥ = 𝑣𝑣𝑣𝑣 cos 𝜙𝜙

𝛷𝛷𝐸𝐸 = 𝐸𝐸�⃗ . 𝐴𝐴⃗ = 𝐸𝐸�⃗ . 𝐴𝐴𝑛𝑛� = 𝐸𝐸𝐸𝐸 cos 𝜙𝜙

𝛷𝛷𝐸𝐸 = 𝐸𝐸�⃗ . 𝐴𝐴⃗ = 𝐸𝐸�⃗ . 𝐴𝐴𝑛𝑛� = 𝐸𝐸(𝜋𝜋𝑟𝑟 2 ) cos 30 = ⋯

Note: direction of vector A is perpendicular to surface; magnitude of vector A equals area of surface If in a plane surface area of ∆𝑠𝑠, a uniform electric field 𝐸𝐸 exists, and makes an angle 𝜃𝜃 with the normal to the surface area (positive normal - you can arbitrarily decide which direction is positive), the quantity ∆𝜙𝜙 = 𝐸𝐸 ∆𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃

is called the flux of the electric field through the chosen surface.∆𝑠𝑠 is represented as a vector. Also it is area. ∆𝛷𝛷 = 𝐸𝐸�⃗ . ∆𝑠𝑠⃗

Where 𝐸𝐸�⃗ and ∆𝑠𝑠⃗ are vectors and ∆𝜙𝜙 is a scalar quantity.

5 As flux is a scalar quantity, it can be added arithmetically. Hence surfaces which are not on single plane, can be divided into small parts which are plane, the flux through each part can be found out and the sum is flux through the complete surface.

Non-uniform electric field can also be tackled that way. Divide the surface into parts over which the electric field is uniform and then find the flux in each part and sum them. Using the techniques of integration flux over a surface is: 𝜙𝜙 = ∫ 𝐸𝐸�⃗ . 𝑑𝑑𝑠𝑠⃗

When integration over a closed surface is done a small circle is placed on the integral sign.(∮ ). Flux over a closed surface

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𝜙𝜙 = ∮ 𝐸𝐸�⃗ . 𝑑𝑑𝑠𝑠⃗ ( Integration over a closed surface) Solid angle:

Typical example is the angle in the paper containers used by moongfaliwalas. 𝛺𝛺 = 𝑆𝑆/𝑟𝑟²

𝑆𝑆 = the area of the part of sphere intercepted by the cone

𝑟𝑟 = radius of the sphere assumed on which we are assuming the cone • • • 3.

A complete circle subtends an angle 2 𝜋𝜋 Any closed surface subtends a solid angle 4 𝜋𝜋 at the centre. The angle subtended by a closed plane curve at an external point is Zero. Gauss's law

The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by 𝜀𝜀0 . In symbols ∮ 𝐸𝐸�⃗ . 𝑑𝑑𝑠𝑠⃗ =

Where, 𝑞𝑞𝑖𝑖𝑛𝑛 = charge enclosed by the closed surface

𝑞𝑞 𝑖𝑖𝑖𝑖 𝜀𝜀 0

𝜀𝜀0 = permittivity of the free space

It needs to be stressed that flux is the resultant of all charges existing in the space. But, its quantity is given the right hand side. 

Some consequences of Gauss's applied to conductors

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A conductor contains at least some charges which are free to move within it. When initially placed in an external E-field the field will penetrate into the conductor and cause the free charges to move (a). However these can only move as far as the surface of the conductor (assuming it is of finite size). Charge collects at the surface and produces an E-field within the conductor which opposes the external field. Equilibrium is quickly reached where the displaced charges produce an internal field which exactly cancels the external one (and hence there is no further movement of charge) (b).

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within a conductor at equilibrium there can be no E-field. all points of the conductor must be at the same potential.

Solid conductor in (a) carries a net charge. Within conductor E=0 hence flux through Gaussian surface G is zero and hence net charge contained within G is also zero. Þ A solid conductor carries all its excess charge on the surface.

Hollow conductor in (b) must also carry any excess charge on its outer surface unless the hollow region contains a charge (+Q) (c) in which case the inner surface must carry an equal but opposite charge -Q. These requirements are necessary to give a zero flux through the Gaussian surface G. It can also be shown that in case (b) E=0 within the hollow region. 4. a.

Applications of Gauss’s law Charged conductor

The free electrons redistribute themselves to make the field zero at all the points inside the conductor. Any charge injected anywhere in the conductor must come over to the surface of the conductor so that the interior is always charge free.

If there is cavity inside the conductor, for example a hollow cylinder and a charge +q is placed in this cavity, as the inside of the conductor has to be charge free, negative charge appears on the inside of the cavity. If the conductor is neutral, a charge +q will appear on the surface.

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= 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 × 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐸𝐸. 4𝜋𝜋𝑟𝑟 2

Total charge enclosed by surface is Q ∴

hence

𝐸𝐸. 4𝜋𝜋𝑟𝑟 2 = 𝑄𝑄/𝜀𝜀0 𝐸𝐸 =

𝑄𝑄

4𝜋𝜋𝜀𝜀 0 𝑟𝑟 2

,

𝑉𝑉 =

𝑄𝑄

4𝜋𝜋𝜀𝜀 0 𝑟𝑟

For r
potential within sphere must be constant and equal to surface potential = b.

Electric field due to a uniformly charged sphere

𝑄𝑄

4𝜋𝜋𝜀𝜀 0 𝑟𝑟

A total charge Q is uniformly distributed in a spherical volume of radius 𝑅𝑅. what is the electric field at a distance r from the centre of the charge distribution outside the sphere? Through Gauss’s law we get So

𝐸𝐸 4 𝜋𝜋𝜋𝜋² =

𝐸𝐸 =

𝑄𝑄

𝑄𝑄

𝜀𝜀 0

4𝜋𝜋𝜀𝜀 0 𝑟𝑟 2

The electric field due to a uniformly charged sphere at a point outside it, is identical with the field due to an equal point charge placed at the centre. Field at an internal point (i.e. 𝒓𝒓 < 𝑹𝑹) At centre 𝐸𝐸 = 𝑂𝑂

At any other point 𝑟𝑟 less than 𝑅𝑅 radius of the sphere 𝐸𝐸 =

𝑄𝑄𝑄𝑄

4𝜋𝜋𝜀𝜀 0 𝑅𝑅 3

Electric field due to a linear charge distribution

The linear charge density (charge per unit length) is 𝜆𝜆.

Electric field at a distance 𝑟𝑟 from the linear charge distribution

c.

𝐸𝐸 =

𝜆𝜆

2𝜋𝜋𝜀𝜀 0 𝑟𝑟

Electric field due to a plane sheet of charge

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Plane sheet with charge density (charge per unit area) σ. Field at distance d from the sheet = 𝐸𝐸 =

𝜎𝜎

2𝜀𝜀 0

We see that the field is uniform and does not depend on the distance from the charge sheet. This is true as long as the sheet is large as compared to its distance from P. d.

Electric field due to a charged conducting surface

To find the field at a point near this surface but outside the surface having charge density σ. 𝐸𝐸 = 𝜎𝜎/ 𝜀𝜀0

Notice in case of plane sheet of charge it is 𝜎𝜎/2𝜀𝜀0 . But in the case of conductor is 𝜎𝜎/ 𝜀𝜀0 . 5.

Spherical Charge Distributions

Useful results for a spherical charge distribution of radius 𝑅𝑅

a) The electric field due to a uniformly charged, thin spherical shell at an external point is the same as that due to an equal point charge placed at the centre of the shell. b) The electric field due to a uniformly charged, thin spherical shell at an internal point is zero. c) The electric field due to a uniformly charged sphere at an external point is the same as that due to an equal point charge placed at the centre of the sphere. d) The electric field due to a uniformly charged sphere at an at an internal point is proportional to the distance of the point from the centre of the sphere. Thus it is zero at the centre and increases linearly as one moves out towards the surface. e) The electric potential due to a uniformly charged, thin spherical shell at an external point is the same f)

as that due to an equal point charge placed at the centre of the shell. 𝑉𝑉 =

𝑄𝑄

4𝜋𝜋𝜀𝜀 0 𝑟𝑟

The electric potential due to a uniformly charged, thin spherical shell at an internal point is the same everywhere and is equal to the that at the surface. 𝑉𝑉 =

𝑄𝑄

4𝜋𝜋𝜀𝜀 0 𝑅𝑅

g) The electric potential due to a uniformly charged sphere at an external point is the same as that due to an equal point charge placed at the centre of the sphere. 𝑉𝑉 =

𝑄𝑄

4𝜋𝜋𝜀𝜀 0 𝑟𝑟

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Electric potential energy of a uniformly charged sphere

Charge density 𝜌𝜌 = 3𝑄𝑄/4𝜋𝜋 𝑅𝑅³ (charge density is per unit volume) Electric potential energy of the charged sphere = 3𝑄𝑄²/20𝜋𝜋𝜀𝜀0 𝑅𝑅

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Electric potential energy of a uniformly charged, thin spherical shell

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Earthing a conductor

Electric potential energy of the thin spherical shell = 𝑄𝑄²/8𝜋𝜋𝜀𝜀0 𝑅𝑅

The earth is good conductor of electricity. Earth’s surface has a negative charge of 1𝑛𝑛𝑛𝑛/𝑚𝑚². All conductors which are not given any external charge are also very nearly at the same potential. But the potential of earth is often taken as zero. When a conductor is connected to earth, the conductor is said to be earthed or grounded and its potential will become zero.

In appliances, the earth wire is connected to the metallic bodies. If by any fault, the live wire touches the metallic body, charge flows to the earth and the potential of the metallic body remains zero. If it not connected to the earth, the user may get an electric shock.