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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
- 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)

- Coulombs (C) - standard

Charges

- Charges can be either
- Negative (-)
- an excess of electrons

- Positive (+)
- a deficiency of electrons

- Negative (-)
- “Fundamental” or “elementary” charge
- Electron = -1.6 x 10-19C
- “Quantized” – integer numbers

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

- 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

- 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

- 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

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?

- 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 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

- 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 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 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”

- Van de Graaff Generator
- Wimshurst Machine

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.

- 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).

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

- 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)

- E = V/d,

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

- Lightning
- Clouds - earth

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