PPU 960 Physics Note [Sem 2 Chapter 12 - Electrostatics]

00

STPM

2013

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Chapter 12 – Electrostatics By : Josh, LRT

2013 © LRT Documents Copyrighted. All rights reserved. Page 1 of 10

Chapter 12 – Electrostatics

Electrostatics is the branch of physics that deals with the phenomena and properties of stationary or

slow-moving (without acceleration) electric charges.

Electric charge

Electrically charged objects have several important characteristics:

Like charges repel one another; that is, positive repels positive and negative repels negative.

Unlike charges attract each another; that is, positive attracts negative.

Charge is conserved. A neutral object has no net charge. If the plastic rod and fur are initially

neutral, when the rod becomes charged by the fur, a negative charge is transferred from the fur

to the rod. The net negative charge on the rod is equal to the net positive charge on the fur.

12.1 Coulomb’s Law

The quantitative expression for the effect of these three variables on electric force is

known as Coulomb's law.

Coulomb's law states that the electrical force between two charged objects is directly

proportional to the product of the quantity of charge on the objects and inversely

proportional to the square of the separation distance between the two objects.

In equation form, Coulomb's law can be stated as:

(

) (

)

Coulomb's Constant

𝑘

≈

9 × 0 𝑁𝑚

𝐶 𝐂𝐨𝐮𝐥𝐨𝐦𝐛 𝐬 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭

The constant of proportionality 𝑘 appearing in Coulomb's law is often called Coulomb's

constant. Note that it can be expressed in terms of another constant ,

× 0 𝑚 .

If the charges q1, q2 are in an insulating medium, then

(

) (

)

Where, 0 , known as the permittivity of the medium, and is a non-dimensional constant

known as the relative permittivity of the medium. *** Insulator .



12.2 Electric Field

Electric field is defined as the electric force per unit charge. The direction of the field is taken

to be the direction of the force it would exert on a positive test charge. The electric field is

radially outward from a positive charge and radially inward a negative point charge.

Electric Field of Point Charge

The electric field of a point charge can be obtained from Coulomb's law:

𝐸

𝑞

𝑘 𝑞

𝑞

𝑘

The electric field is radially outward from the point charge in all directions. The circles

represent spherical equipotential surfaces.

The electric field from any number of point charges can be obtained from a vector sum of the

individual fields. A positive number is taken to be an outward field; the field of a negative

charge is toward it.

Electric Field Intensity

Electric field strength is a vector quantity; it has both magnitude and direction.

The magnitude of the electric field strength is defined in terms of how it is measured.

The test charge has a quantity of charge denoted by the symbol q. When placed within the

electric field, the test charge will experience an electric force - either attractive or repulsive.

If the electric field strength is denoted by the symbol E, then the equation can be rewritten

in symbolic form as:

𝐸

, 𝑢𝑛𝑖𝑡 𝑵𝑪



Multiple Point Charges

The electric field from multiple point charges can be obtained by taking the vector sum of the

electric fields of the individual charges.

After calculating the individual point charge fields, their components must be found and added

to form the components of the resultant field. The resultant electric field can then be put into

polar form. Care must be taken to establish the correct quadrant for the angle because of

ambiguities in the arctangent.

Motion of a charge in a Uniform Electric Field

For a given separation and potential differences, the electric field strength

between two parallel plates is constant (uniform field), except at the

edges. A point charge q is moved between the plates and experiences a

constant force F due to the uniform electric field.

• Work done on the charge = energy transformed by the charge

[ 𝐹𝑑 𝑉𝑞 ]

• Rearranging this equation, we have [ 𝐹

𝑞

𝑉

𝑑 ]

• E is directly proportional to the potential differences between the two

plates and inversely proportional to the separation of the plates.

[𝐸 𝑉

𝑑 ]

• Units of electric field strength are 𝑉𝑚 or 𝑁𝐶 .



12.3 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 electric flux through an area is defined as the electric field multiplied by the area of the

surface projected in a plane perpendicular to the field.

Gauss's Law is a general law applying to any closed surface. It is an important tool since it

permits the assessment of the amount of enclosed charge by mapping the field on a surface

outside the charge distribution.

If it picks any closed surface and steps over that surface, measuring the perpendicular field

times its area, it will obtain a measure of the net electric charge within the surface, no matter

how that internal charge is configured.

The general equation is 0

.

Electric Flux

The concept of electric flux is useful in association with Gauss' law. The electric flux through

a planar area is defined as the electric field times the component of the area perpendicular to

the field. If the area is not planar, then the evaluation of the flux generally requires an area

integral since the angle will be continually changing.

When the area A is used in a vector operation like this, it is understood that the magnitude of

the vector is equal to the area and the direction of the vector is perpendicular to the area.

The general equation is 𝐸 .



Gaussian Surfaces

Part of the power of Gauss' law in evaluating electric fields is that it applies to any surface. It is

often convenient to construct an imaginary surface called a Gaussian surface to take advantage of

the symmetry of the physical situation.

If the symmetry is such that you can find a surface on which the electric field is constant, then

evaluating the electric flux can be done by just multiplying the value of the field times the area of

the Gaussian surface.

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. The electric flux is then just the electric

field times the area of the sphere.

𝐸 𝐸

0

The electric field at radius is then given by: 𝐸

0

If another charge is placed at , it would experience a force: 𝑞

0

This is seen to be consistent with Coulomb's law.



Electric Field of Conducting Sphere

The electric field of a conducting sphere with charge Q can be obtained by a

straightforward application of Gauss' law. Considering a Gaussian surface

in the form of a sphere at radius , the electric field has the same

magnitude at every point of the surface and is directed outward. The electric

flux is then just the electric field times the area of the spherical surface.

𝐸 𝐸

0

The electric field is seen to be identical to that of a point charge Q at the center of the sphere.

Since all the charge will reside on the conducting surface, a Gaussian surface at will

enclose no charge, and by its symmetry can be seen to be zero at all points inside the spherical

conductor.

For ,

𝐸

For ,

𝐸 0

12.4 Electric Potential

The total of the electric flux out of a closed surface is equal to the charge enclosed divided by

the permittivity.

Potential: Charged Conducting Sphere

The use of Gauss' law to examine the electric field of a charged sphere shows that the electric

field environment outside the sphere is identical to that of a point charge. Therefore the

potential is the same as that of a point charge:

𝑉 𝑘

𝑉 𝐸 𝑘

The electric field inside a conducting sphere is zero, so

the potential remains constant at the value it reaches at

the surface:



Potential: Potential of a Conductor

When a conductor is at equilibrium, the electric field inside it is constrained to be zero.

Potential Difference

The equation connecting work , charge and potential difference 𝑉 is as follows:

𝑉

The diagram below illustrates that the work done in taking charge around a closed loop is zero.

Work W is done by the electric field in moving the charge from 𝑉 to . However, work must

be done against the field to return the charge back to 𝑉 .

So the sum amount of 0 paths and work done.

Since the electric field is equal to the rate of change of

potential, this implies that the voltage inside a

conductor at equilibrium is constrained to be constant

at the value it reaches at the surface of the conductor.

A good example is the charged conducting sphere,

but the principle applies to all conductors at

equilibrium.



Relation between E and V

Consider a charge being moved by a force F from an arbitrary point A to another point B

against an electric field of strength .

The distance moved, is very small, such that the force F may be considered constant.

Hence the work done by the force is:

The force is equal to the force exerted by the field on the charge, but in the opposite direction.

𝐸

Substituting in the original equation for gives :

𝐸

From the definition of potential difference,

Therefore, if the potential difference between A & B is and ( )

𝑉

Substituting for ,

𝐸 𝑉

Cancelling the Q and rearranging,

𝐸 𝑉

In the limit as and tend to zero,

𝑉

𝑉



Multiplying both sides by :

𝐸 𝑉

Equation for graph sketching:

𝐸 𝑉

The E-r and V-r graphs below show the relation clearly.

The gradient of the V-r graph is negative. So the negative of its gradient gives a positive value for

E in the E-r graph.



Problems

1. Where the field strength is 000 𝑁𝐶 , what is the force on a 𝐶 charge on an electron?

2. A charged sphere is placed in a field of strength 0 𝑁 𝐶 . If it experiences a force of

𝑁, what is the charge on the sphere?

3. What is the field strength if an electron experiences a force of 0 𝑁?

4. A charged dust particle is stationary between two horizontal charged metal plates. The metal

plates have a separation of 𝑚 and the potential difference between the plates is 0 𝑉. The

dust particle has a charge of . Calculate:

i) The electric field strength between the plates

ii) The weight of the dust particle

5. Diagram below shows the variation of electric field with distance from a point charge.

6. Which of the following statements is true for an isolated hollow sphere conductor which has a

fixed negative charge?

7. The work done to bring two point charges 𝑞 and 𝑞 from infinity to separation is

The shaded area below the graph between P and Q represents

A. Work done to bring a unit charge from P to Q.

B. Electric field intensity between P and Q.

C. Total charge that moves from P and Q.

D. Potential difference between P and Q.

A. Electric field strength outside the sphere is less than the electric field strength inside the

sphere.

B. Potential gradient at points outside the sphere is independent of the size of the sphere.

C. Electric field outside the sphere is radial and pointing outwards from the center of the

sphere.

D. Electric potential on the surface of the sphere is inversely proportional to the square of the

radius of the sphere.

A. 𝑞

𝜋𝜀 𝑟

B. – 𝑞

𝜋𝜀 𝑟

C. 𝑞

𝜋𝜀 𝑟

D. – 𝑞

𝜋𝜀 𝑟

Answers:

1. 1.6 x 10-16

N

2. 5 x 10-4

C

3. 3 x 105 N C

-1

4. i) 2.0 x 104 V m

-1 , ii) 2.2 x 10

-14 N

5. D

6. B

7. B