Electrostatic Force and Electric Fields - Learn

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 2^nd 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 × 10⁹Nm²C⁻². This is usually rounded off to 9.0 × 10⁹Nm²C⁻² for calculations.

Coulomb’s Law can now become:

$\vec { F } =k\cfrac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } }$

where k = 9.0 × 10⁹Nm²C⁻²

Example 1:

Calculate the uniform electric field that will cause a force of 2.4 × 10⁻²² N on an electron (q_e = −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 (q_p = 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)