Electrostatic Force and Electric Fields – Learn
Electric field strength allows us to quantitatively look at electric fields. Mathematically, the electric field, \vec { E } , is given by:
\vec { E } =\cfrac { \vec { F } }{ q }
where;
\vec { E } = the electric field strength in { NC }^{ -1 }
\vec { F } = the force on the charged particle in N
{ q } = the charge on the object experiencing the force in C
The previous equation is also commonly arranged to give:
\vec { F } =q\vec { E }
The acceleration of a charged particle in an electric field can also be determined by using Newton’s 2nd Law:
\vec { F } =m\vec { a }
where;
\vec { a } = the acceleration of the charged particle in { m }/{ { s }^{ 2 } }
m = the mass of the accelerating particle in kg
Coulomb’s Law
Coulomb’s Law is used to determine the force between two charged particles:
\vec { F } =\cfrac { 1 }{ { 4\pi \varepsilon }_{ 0 } } \times \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }
where:
\vec { F } = the force on each charged object (in N)
{ q }_{ 1 } and { q }_{ 2 } = the charges on the two points (in C)
r = the distance between each charged point (in m)
{ \varepsilon }_{ 0 } = the permittivity of free space, which is equal to 8.854 × 10−12 A2S4Kg−1m−3 in air or in a vacuum.
\cfrac { 1 }{ { 4\pi \varepsilon }_{ 0 } } is known as Coulomb’s constant, k, and is equal to 8.9875 × 109 Nm2C−2. This is usually rounded off to 9.0 × 109 Nm2C−2 for calculations.
Coulomb’s Law can now become:
\vec { F } =k\cfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }
where k = 9.0 × 109 Nm2C−2
Example 1:
Calculate the uniform electric field that will cause a force of 2.4 × 10−22 N on an electron (qe = −1.602 × 10−19 C)
Answer:
using: \vec { E } =\cfrac { \vec { F } }{ q }
\vec { E } =\cfrac { { 2.4\: \times \: 10 }^{ -22 } }{ { -1.602\: \times \: 10 }^{ -19 } }
\vec { E } =-0.0015\: { NC }^{ -1 } , or
\vec { E } =0.0015\: { NC }^{ -1 } in the opposite direction to the force
Example 2:
Calculate the acceleration of the electron in example 1 (mass of an electron = 9.1 × 10−31 Kg):
Answer:
using:
\vec { F } =m\vec { a } and rearranging to give:
\vec { a } =\cfrac { \vec { F } }{ m }
\vec { a } =\cfrac { { 2.4\: \times \:10 }^{ -22 } }{ { 9.1\: \times \: 10 }^{ -31 } }
\vec { a } =\: { 2.64\: \times \: 10 }^{ 8 }\: { m }/{ { s }^{ 2 } }
Example 3:
Calculate the force between 2 protons that are separated by 5mm (qp = 1.602 × 10−19 C):
Answer:
using: \vec { F } =\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }
\vec { F } =\cfrac { 1 }{ 4\pi { (8.854\times { 10 }^{ -12 }) } } \cfrac { (1.602×{ 10 }^{ -19 })(1.602×{ 10 }^{ -19 }) }{ { 0.005 }^{ 2 }}
\vec { F } =9.24\times { 10 }^{ -24 }\: N (repulsion)