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Electrostaticfields

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Introduction

Applications:

Cathode ray tubes

Devices: Keyboards, touchpads, LCDs

Machines: X-ray machines,Electrocardigrams

Industrial processes: spraypainting, electrodeposition,electrostatic separation ofsolids

An electrostatic field is produced by a static chargedistribution.

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Electrostatic Force

ANY charged object can exert the electrostatic forceupon other objects- both charged and uncharged

objects.

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Coulombs Law- formula for

electrostatic force

this is similar to the gravitational force

Fg= GmM

r2charge (q) is now

responsible for the forceFe= kq1q2

r2

Just like G was a constant so is k.

k is the electrostatic constant

and = 8.99 x 109Nm2/C2

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Vector form of Coulombslaw

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Problem 1:

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Electric field due to continuous chargedistribution

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Electric field due to infinite line charge

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Homework : Infinite Surface Charge

El t i Fl

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Electric Flux

Lines of force in any particular electric field

is called the electric flux. The flux lines start from positive charge andterminate on the negative charge.

If one type of charges is absent?

More number of flux lines? Do they cross each other?

We define electric flux in terms of electric

flux density (D)

The relation between Dand E:

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Gauss Law: Maxwells

first equation

Gauss's law states that the total electric flux through any closed surface is equal to the totalcharge enclosed by that surface. Thus

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Applications of Gausss law The Gauss law can be used to find electric field

for symmetrical charge distribution. Gaussslaw can not be used to find electric field if

charge distribution is not symmetric; we mustresort to Coulombslaw to determine electric fieldin that case.

The calculation would be extremely difficultbecause the electric field term occurs inside theintegral in the Gausslaw equation.

However, in cases that have a high degree ofsymmetry, it is possible to choose a Gaussiansurface such that the electric field term can bemoved outside of the integral.

C diti f l ti th

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Condition for selecting theclosed Gaussian surface

Ds is everywhere either normal ortangential to the Gaussian surface, sothat Ds.dSbecomes either DsdS orzero, respectively.

On the portion of the closed surface forwhich Ds.dS is not zero, Ds= Constant.

This allows us to replace the dotproduct with DsdS and then to bring Dsoutside the integral sign.

Must know direction of electric fieldfrom symmetry of problem.

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There are three such symmetries forwhich this procedure is possible,corresponding to our three spatial

dimensions.

We have the full three-dimensionalsymmetry of the sphere, the two-

dimensional symmetry of the infinitecylinder, and the one-dimensionalsymmetry of the infinite plane.

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Infinite Line Charge:

r

Suppose the infinite line of uniform charge LC/m lies along the z-

axis. To determine Dat a pointP, we choose a cylindrical surfacecontainingPto satisfy symmetry condition as shown in Figure. Dis

constant on and normal to the cylindrical Gaussian surface. If we

apply Gauss's law to an arbitrary lengthl of the line

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Point Charge

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Problem 3: Finite Line Charge

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Electric Potential

In determining VAB, A is the initial point whileB is the final point.

If VABis negative, there is a loss in potential energy in moving Q fromA toB;

this implies that the work is being done by the field. However, if VAB ispositive, there is a gain in potential energy in the movement; an external agent

performs the work.

VABis independent of the path taken

VABis measured in joules per coulomb, commonly referred to as volts (V).

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Problem 2:

Consider an infinite line charge alongz-axis. Show that the work done iszero if a point charge Q is moving in

a circular path of radius r, centeredat the line charge.

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Concept of absolute potential

if the Efield is due to a point

charge Q located at the origin,

then

where VB and VA are the absolute potentials

atBandA, respectively.

Absolute potentials are measured w.r.t.a specified reference position. Such areference position is assumed to be atzero potential. Here we assume thepotential at infinity is zero and hence,we choose infinity as reference. Thus

the absolute potential at any point rdue to a point charge Q located at theorigin is

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If the point charge Q is not located at the origin but at

a point whose position vector is r', the potential V(x, y,

z) or simply V(r) at rbecomes

For n point charges Q1Q2,. , Qnlocated at points with position

vectors r1, r2,. . ., rn, the potential at ris

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Equipotential surfaces

Conservative Field

Relation between Eand V: Energy density in electrostatic field