Electrostatics
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Electrostatics. Overview. Electrostatics - Concepts. Definition Electrostatics = Electron + stationary “The study of electric charges , forces and fields” Symbol for charge Letter “q” or “Q” Units of charge Coulombs (C) - standard

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Electrostatics

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Electrostatics

Electrostatics

Overview


Electrostatics concepts

Electrostatics - Concepts

  • Definition

    • Electrostatics = Electron + stationary

    • “The study of electric charges, forces and fields”

  • Symbol for charge

    • Letter “q” or “Q”

  • Units of charge

    • Coulombs (C) - standard

      • 1 C = the charge on 6.25 x 1018 electrons, therefore

      • 1 electron has a charge of 1.6 x 10-19 C

    • microCoulombs (µC) - common

      • X 10-6 (ex. 3 x 10-6 C = 3 µC)


Charges

Charges

  • Charges can be either

    • Negative (-)

      • an excess of electrons

    • Positive (+)

      • a deficiency of electrons

  • “Fundamental” or “elementary” charge

    • Electron = -1.6 x 10-19C

    • “Quantized” – integer numbers


Atomic structure

Atomic Structure

  • Electrons are free to move, but not protons or neutrons

  • Positive and negative particles, when brought close together are subject to the following “BIG RULE”:

    • Like (+/+ or -/-) charges repel

    • Unlike (+/-) charges attract

Overall – charge is conserved


Electrostatic force coulomb s law

Electrostatic Force - Coulomb’s Law

  • Two charged particles q1 & q2 interact with each other with a force that is:

    • Proportional to the product of their charges, and

    • Inversely proportional to the square of their separation distance (d)

    • where

    • k = electrostatic constant

    • k = 9 x 109 N.m2/C2

  • (Cf Newton’s Law of Universal Gravitation)


Practice coulombs law

Practice – Coulombs’ Law

  • What is the magnitude of the e-static force between an electron and proton of hydrogen, when they are separated 5.3 x 10-11 m? (qe = -1.6 x 10-19 C, qp = 1.6 x 10-19 C)

  • Solve Coulomb’s Law for F

    • F = k(q1q2)/d2

    • F = 9 x 109 x (-1.6 x 10-19 C) x (1.6 x 10-19 C)

      (5.3 x 10-11)2

    • F = 8.2 x 10-8 N


Practice coulomb s law

Practice – Coulomb’s Law

  • Two e-static charges of 60 x 10-6C and 50 x 10-6C exert a force of 175N on each other. What is their separation distance?

  • Solve Coulomb’s Law for d:

    • F = k(q1q2)/d2

    • 175 = 9 x 109 x (60 x 10-6)(50 x 10-6)

      d2

    • d2 = 1.54 x 10-1

    • d = 3.93 x 10-1 m


Some results of e static charges

Some results of E/static charges


Electric field e

Electric Field (E)

  • An electric field is a region in space in which an electric charge experiences a force.

  • Surrounding electric charges are lines of force which form an electric field. These lines of force flow out of a positive charge and into a negative charge, as shown.

  • Represented by symbol E

  • E is a vector!

E


Electric field strength

Electric Field Strength?

  • The Electric Field strength can be determined by 2 methods:

  • Method 1: The strength of this electric field (E) at a point is equal to the force felt by a small positive test charge (q0) being placed at that point.

    • E = F/q0 Newtons/Coulomb (N/C)


Electric field strength1

Electric Field Strength?

  • Method 2: the electric field strength at a point P, d units distant from the charge of interest q is equal to the product of the charge and the electrostatic constant (k) divided by the square of the separation distance, or

    • E = kq/d2 (N/C) 


Practice electric field

Practice – Electric Field

  • A force of 2.1 N is exerted on a 9.2e-4 C test charge when it is placed in an electric field created by a 7.5 C charge. If the force is pushing it West,

    • determine the electric field (E) at that point.

  • Solve E = F/q0 for E

    • E = 2.1/9.2e-4

    • E = 2,282.6 N/C


Practice electric field1

Practice – Electric Field

  • If a positive test charge of 3.7e-6 C is put in the same place in the electric field as the original test charge in the last example,

    • determine what force (F) will be exerted on it.

  • Solve E = F/qo for F

    • 2282.6 = F/3.7e-6

    • F = 8.44e-3 N


Practice electric field2

Practice – Electric Field

  • A tiny metal ball has a charge of –3.0e-6 C.

    • determine the magnitude and direction of the field at a point, P, 30cm away?

  • Solve E = kq/d2 for E

    • E = (9e9)(-3e-6)/0.32

    • E = -300,000 N/C


Electrostatic toys

Electrostatic “Toys”

  • Van de Graaff Generator

  • Wimshurst Machine


Electrical potential difference voltage v

Electrical Potential Difference (Voltage V)

  • Definition

    • The amount of work done (change in electrical potential energy) in moving a +ve test charge (q0)from one location to another in an electrical field (E).

      • V = W/q0 in joules/coulomb = volts

    • If the charge is moving against the field, then the work done is negative, and vice versa.


Practice potential difference

Practice – Potential Difference

  • Determine the electric potential difference (or voltage V) of a 3.4 C charged object (q) that gains 2.6 x 103 J as it moves through an electric field.

  • Solve V =W/q for V

    • V = (2.6x 103)/3.4C

    • V = 7.6 x 102V


Parallel plates

Parallel Plates

  • If an electric field (E) exists between 2 parallel plates, then the field is uniform except at the edges.

  • The electric field depends on the potential difference (V) across the plates and the plate separation (d), as follows:

    • E = V/d,

      • units of volt/meter (equivalent to N/C)


Practice parallel plates

Practice – Parallel Plates

  • If we have two parallel plates that are 16.0mm apart and want a uniform field of 800 N/C between these plates, determine the voltage we must apply to the plates.

  • Solve E = V/d for V

    • V = (800N/C) (0.0160m)

    • V = 12.8 V


Example of parallel plates

Example of Parallel Plates

  • Lightning

    • Clouds - earth


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