Electrostatics

The Electric Force

 Electrostatics

      The Electric Force

           Coulomb's Law

Phenomenological approach

Summary of evidence from observations:

*Like charges repel, unlike attract

For point charges:

*Changing either charge changes the force proportionally. (e.g. doubling either charge doubles the force) 

*The force decreases as the distance between the charges increase, and vice versa.

*The force changes quadratically with the distance (e.g. halving the distance quadruples the force) 

Electrostatics

The Electric Charge

... and the rest of the cast

 Electrostatics

      The influence & interaction of electric charges

           The Cast 

q
\vec{E}
V
\Phi
\vec{F}
U

potential

potential energy

field

force

charge

flux

influence

interaction

Electric ....

 Electrostatics

      The influence & interaction of electric charges

           The Cast - relationship map

Electric ....

\vec{E}
V
\Phi

influence

interaction

\ \vec{E}=-\vec{\nabla} V\
\ \Delta V =- \int \vec{E}\cdot d\vec{s}\
\ \Phi = \int \vec{E}\cdot d\vec{A}\
\vec{F}
U
\ \vec{F}=-\vec{\nabla} U\
\ \Delta U = -\int \vec{F}\cdot d\vec{s}\
\ \vec{F} = q_0\ \vec{E}\
\ \vec{E} = \vec{F}/q_0\ \
\ U = q_0\ V\
\ V = U/q_0\ \
q

Electrostatics

The Electric Field

and The Electric Force

 Electrostatics

      The Electric Field

           Relationship to the Electric Force

\vec{E}

influence at some location in space

interaction between charges

\vec{F}
\ \vec{F} = q_0\ \vec{E}\
\ \vec{E} = \vec{F}/q_0\ \

Electric Field

Electric Force

 Electrostatics

      The Electric Field

           Relationship to the Electric Force

\ \vec{F} = q_0\ \vec{E}\
\vec{E}

influence at some location in space

interaction between charges

\vec{F}
\ \vec{F} = q_0\ \vec{E}\
\ \vec{E} = \vec{F}/q_0\ \

Electric Field

Electric Force

\text{a charge $q_0$ in a local field $\vec{E}$}
\text{experiences a force $\vec{F}=q_0\vec{E}$}
\vec{E}
\vec{E}
\vec{E}
\vec{F}
\text{the force on a positive charge}
\text{is in the same direction as the field}
\text{the force on a negative charge}
\text{is in the opposite direction to the field}
\vec{F}

Electrostatics

The Electric Force

between two point-charges

 Electrostatics

      The Electric Force

           between two point charges

q_1
q_2
\vec{E}_\text{@ P due to $q_1$}=\frac{kq_1}{r^2}\ \hat{r}

The Electric Field generated by q1 at the location P:

Another charge q2 placed at P would experience a force: 

\vec{F}_\text{on a charge $q_2$ present @P}= q_2\times \vec{E}_\text{@ P due to $q_1$}
\vec{F}_\text{$q_1$ on $q_2$}=\frac{kq_1\ q_2}{r^2}\ \hat{r}_{\tiny q_1\to q_2}

Putting it together:

 Electrostatics

      The Electric Force

           Coulomb's Law

\vec{F}_{ij}=\tfrac{1}{4\pi\epsilon_0}\frac{q_i\ q_j}{r^2_{\tiny ij}}\ \hat{r}_{\tiny ij}
\text{the electric force of $q_i$ on $q_j$}
\text{permittivity}
\text{ the distance between $q_i$ and $q_j$}
\text{ direction $i\rightarrow j$}
\text{ square}
\text{the charge exerting the force}
\text{the charge experiencing the force}
\left( k =\frac{1}{4\pi\epsilon_0} =8.99\times10^9 \quad \frac{\text{N$\cdot$ m$^2$}}{\text{C}^2}\right)
\epsilon_0 =8.85\times10^{-12} \quad \frac{\text{C}^2}{\text{N$\cdot$ m$^2$}}
\text{(free-space)}
\vec{F}_{q_i \text{ on } q_j}

 Electrostatics

      The Electric Force

           Coulomb's Law -- direction information

\vec{F}_{ij}=k\frac{q_i\ q_j}{r^2_{\tiny ij}}\ \blue{\hat{r}_{\tiny ij}}
\textcircled{\mathbf{\cdot}}
\textcircled{\cdot}
q_j
q_i
\vec{r}_{i}
\vec{r}_{j}
\text{ $\vec{r}_{ij}$ originates at $i$ and terminates at $j$}
\vec{r}_{ij}= \vec{r}_{j}-\vec{r}_{i}
\hat{r}_{ij}\text{ is a unit vector in direction of $\vec{r}_{ij}$}
\hat{r}_{ij}=\tfrac{\vec{r}_{ij}}{r_{ij}}
\hat{r}_{ji}= -\hat{r}_{ij}
\vec{r}_{ji} = \vec{r}_{i}-\vec{r}_{j} =-( \vec{r}_{j}-\vec{r}_{i}) =-\vec{r}_{ij}
\vec{r}_{ji}\text{ is in a direction opposite of of $\vec{r}_{ij}$}
\vec{F}_{ji} = k\frac{q_j\ q_i}{r^2_{\tiny ji}}\ \blue{\hat{r}_{\tiny ji}} = k\frac{q_i\ q_j}{r^2_{\tiny ij}}\ \blue{(-\hat{r}_{\tiny ij})} =-\vec{F}_{ij}
\hat{r}_{ij}
\vec{r}_{ij}
q_i
\textcircled{\mathbf{\cdot}}
\vec{r}_{ji}
\hat{r}_{ji}
\vec{r}_{i}
\textcircled{\cdot}
q_j
\vec{r}_{j}

 Electrostatics

      The Electric Force

           Coulomb's Law -- attraction & repulsion

q_j
q_i
\vec{F}_{ij}=k\frac{\blue{q_i\ q_j}}{r^2_{\tiny ij}}\ {\hat{r}_{\tiny ij}}
q_i\ q_j\gt 0
q_i\ q_j\lt 0
q_j
q_i
q_j
q_i
q_i
q_j
\vec{F}_{ij}
\vec{F}_{ij}
\vec{F}_{ij}
\vec{F}_{ij}
\hat{r}_{ij}
\hat{r}_{ij}
\hat{r}_{ij}
\hat{r}_{ij}
q_j
q_i
\hat{r}_{ij}
\vec{r}_{ij}

Electrostatics

The Electric Force

The net Electric Force due to a configuration of charges

 Electrostatics

      The Electric Force

           Coulomb's Law

\textcircled{\mathbf{+}}
\textcircled{-}
\textcircled{+}
\textcircled{-}

For a given configuration of point charges

q_1
q_2
q_4
q_3

 Electrostatics

      The Electric Force

           Coulomb's Law

\textcircled{\mathbf{+}}
q_1
\textcircled{-}
\textcircled{+}
q_2
\textcircled{-}
q_4
q_3

The interaction can be described in terms of force-pairs

\text{the electric force}
\text{of $q_4$ on $q_2$}
\text{the electric force}
\text{of $q_2$ on $q_4$}

 Electrostatics

      The Electric Force

           Coulomb's Law

\textcircled{-}
q_4

The net electric force on a charge of interest is the vector sum of all the electric forces acting on it.

\text{the electric force}
\text{of $q_2$ on $q_4$}
\textcircled{-}
q_4
\text{the electric force}
\text{the electric force}
\text{of $q_1$ on $q_4$}
\Sigma\vec{F}_\text{on $q_4$}
\vec{F}_\text{$q_1$ on $q_4$}
\vec{F}_\text{$q_2$ on $q_4$}
\vec{F}_\text{$q_3$ on $q_4$}
=
+
+
\text{of $q_3$ on $q_4$}
\Sigma \vec{F}_\text{on $q_4$}
\vec{F}_\text{$q_1$ on $q_4$}
\vec{F}_\text{$q_2$ on $q_4$}
\vec{F}_\text{$q_3$ on $q_4$}

 Electrostatics

      The Electric Force

           Coulomb's Law -- Check your understanding

Three identical small spheres are fixed on the vertices of an equilateral triangle whose side is [d] cm in length. The spheres at vertices A and B carry negative excess charges, qA and qB, respectively. The sphere at vertex C carries excess positive charge, qC.

 

Suppose that qA=qB, what is the magnitude and direction of the net electric force on the sphere at C due to the charges at A and B.

The force on C due to the combined effect of A and B is the vector sum of the forces on C due to A, and on C due to B:

\vec{F}_\text{on C due to A and B} = \vec{F}_\text{on C due to A} + \vec{F}_\text{on C due to B}

The magnitude of each of the forces is given by Coulomb's law. However, since these forces are pointing in different directions, the only way to find their resultant is by vector addition:

Due to the symmetry, the x components of the forces are going to cancel each other because they would be equal and opposite. The y components would add:

(F_\text{on C due to A and B})_x = - \frac{k\ q_A\ q_C}{r^2_{AC}} \cos{60^\circ } + \frac{k\ q_B\ q_C}{r^2_{BC}} \cos{60^\circ } = - \frac{k\ q\ q_C}{d^2} \cos{60^\circ } + \frac{k\ q\ q_C}{d^2} \cos{60^\circ }=0
(F_\text{on C due to A and B})_y = - \frac{k\ q_A\ q_C}{r^2_{AC}} \sin{60^\circ } - \frac{k\ q_B\ q_C}{r^2_{BC}} \sin{60^\circ } = - \frac{k\ q\ q_C}{d^2} \sin{60^\circ } - \frac{k\ q\ q_C}{d^2} \sin{60^\circ }=- 2\ \frac{k\ q\ q_C}{d^2} \sin{60^\circ }

 Electrostatics

      The Electric Force

           Coulomb's Law -- Check your understanding

\textcircled{\mathbf{+}}
\textcircled{-}
\textcircled{+}
\textcircled{-}
q_1
q_2
q_4
q_3