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Notes 25 - Chapter 16 – Electric Fields
CP030812A
See “Electrostatic” Handout
Electric Field: Every ion (charged particle) has an electric field surrounding it.
The symbol we use to determine the strength of an electric field is _____
The units for the strength of an electric field are ___________________.
For an ion with a position charge The electric field cab be considered to look like this(moving away from the ion)
E
E for a positive ion is ALWAYS +
E
For an ion with a negative charge, the electric field cab be considered to look like this(moving into the ion)
E for a negative ion is ALWAYS -
The attractive force between unlike charged ions can be explained by this diagram
The repulsive force between like charges can be explained by these diagrams.
Like Positive Charges
Like Negative Charges
The strength of the electric field produced by the charge on an ion at a certain distance
from the ion, can be expressed using this formula-
E = Electric Field Strength
(units n/C)
k = The electrostatic constant = 9 x 109 units = n m2/C2
This value never changes. It is constant in all problems
Q = The charge on the ion
R = The distance from the ion
Example Problem 1: An ion has a charge of + 2.4 x 10-18 C. What is the strength
of the electric field at a distance of 6 x 10-6 m from the ion?
E= kQ
R2
E = 9 x 109 (2.4 x 10-18)
(6 x 10-6)2
E= + 600 n/C
NOTE: The + means that the charge on the ion is positive.
Example Problem 2: An ion has a charge of - 9.6 x 10-17 C. What is the strength
of the electric field at a distance of 5 m from the ion?
If another ion is brought into an electric field, it will experience an attractive or
repulsive force.
Diagram : Q1 is the + charged ion producing a + electric field
Q2 is another ion, only – charged brought into the electric field
Hence, there is a resultant attractive force
The equation for this force is
FE = electro static force (units n)
E = is the electric field strength (unit n/C)
Q = the charge on the ion brough into this electric field (unit C)
Example Problem 1:
An ion with a charge of 6.4 x 10-5 C is brought into an electric field with a strength
of 9 x 107 n/C. What electrostatic force will it experience?
FE = E Q
FE = (9 x 107) (6.4 x 10-5)
FE = 5760 n
Example Problem 2An ion with a charge of 3.2 x 10-6 C is brought into an electric field with a strength
of 6 x 105 n/C . What is the electrostatic force on the ion?
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