Electrostatics

The influence of electric charges

The Electric Field

 Electrostatics

      The Electric Field

           Learning Outcomes

Learn how to:

Calculate the electric field at a given point in space due to

  • a configuration of point charges.
  • a "continuous" distribution of electric charges.

Relate the electric field to

  • the electric potential
  • the electric 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 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\quad
\ V = U/q_0\ \
q
\Phi_g

influence

\ \Phi_g = \int \vec{g}\cdot d\vec{A}\
\vec{g}
gy
\ \vec{g}=-\vec{\nabla} (gy)\
\ \Delta V = \int \vec{g}\cdot d\vec{s}\
\vec{F}_g
mgy

interaction

\ \vec{F}=-\vec{\nabla} (mgy)\\ =-mg\hat j
\ \Delta U = -\int \vec{F}\cdot d\vec{s}\
m
\ \vec{F_g} = m\ \vec{g}\
\ \vec{g} = \vec{F_g}/m\ \
\ U_g = m\ V_g\
\ V_g = U_g/m\ \

gravitational....

analogus to

Electrostatics

The Electric Charge

Charge Distribution

 Electrostatics

      Electric Charge

           Charge Distributions

\lambda = dq / dl

Linear Charge Density

\sigma=dq / dA

Surface Charge Density

\rho=dq / dV

Volume Charge Density

\lambda_\text{\tiny uniform} = \frac{Q_\text{\tiny total}}{l_\text{\tiny total}}
\sigma\text{\tiny uniform} = \frac{Q_\text{\tiny total}}{A_\text{\tiny total}}
\rho\text{\tiny uniform} = \frac{Q_\text{\tiny total}}{V_\text{\tiny total}}

 Electrostatics

      Electric Charge

           Point Charge

Charge distribution A

Charge distribution B

What do we mean by point charges?

\text{size$_\text{charge distribution}$}
\text{scale of analysis}
\lt\lt\lt

Electrostatics

The Electric Field

due to a point charge

 Electrostatics

      The Electric Field

           due to a point charge

The Electric Field due to a point charge is    

\vec{E}(r)=\frac{kq}{r^2}\hat{r}

given by ...

charge creating the field.

distance from the charge q to the point of interest P

unit-vector pointing away from the charge towards P

\hat{r}:
q:
r:

 Electrostatics

      The Electric Field

           due to a point charge

\vec{E}(r)=\frac{kq}{r^2}\hat{r}

charge creating the field.

distance from the charge q to the point of interest P

unit-vector pointing away from the charge towards P

\hat{r}:
q:
r:
+
-
\vec{E}
\vec{E}

 Electrostatics

      The Electric Field

           due to a point charge

\vec{E}(r)=\frac{kq}{r^2}\hat{r}

charge creating the field.

distance from the charge q to the point of interest P

unit-vector pointing away from the charge towards P

\hat{r}:
q:
r:
+
\vec{E}
\vec{E}
-

 Electrostatics

      The Electric Field

           Example

q=+1.0\ \text{nC}

Find the magnitude and direction of the electric field.

 Electrostatics

      The Electric Field

           Example

q=-2.0\ \text{nC}

Find the magnitude and direction of the electric field.

 Electrostatics

      The Electric Field

           Example

Find the magnitude of the charge generating the field.

 Electrostatics

      The Electric Field

           Example

q=-4.0\ \text{nC}

Find the magnitude and direction of the electric field.

Electrostatics

The Electric Field

due to a configuration of point charges

 Electrostatics

      The Electric Field

           Multiple point charges

\vec{E}_\text{@ P}={\Large \Sigma}_{\tiny i} ^{\tiny N} \ \frac{k\ q_i}{r^2_i}\hat{r}_i
\text{The Net Electric Field due to multiple point-charges }
\text{vector sum}
\text{the contribution due to $q_i$}

Electrostatics

The Electric Field

Representation and Visualization

 Electrostatics

      The Electric Field

           Visualizing the Electric Field

Visualizing the Electric Field from point charges -- phet

In this simulation, the electric field is represented on a grid by arrows whose color indicates their magnitude.

 

The yellow sensors detect the magnitude and direction of the electric field at any given point in space.

 Electrostatics

      The Electric field

           Visualizing the Electric Field

Visualizing the Electric Field

In this simulation, you can switch between presenting the field as vectors (arrows) or by electric field lines (EFLs).

The electric field lines follow the vectors in tip-to-tail sequence.

Electrostatics

The Electric Field

The Electric Field Lines

 Electrostatics

      The Electric Field

           Electric Field Lines and Equipotential Surfaces

The "landscape" analogy

The electric field at any point is the steepest slope of the electric potential -- if the potential is smooth, the steepest slope is a continuous curve.

If you follow that curve, you are traveling on an electric field line.

Electric Field lines are perpendicular to the equipotential surfaces.

 Electrostatics

      The Electric field

           Visualizing the Electric Field

Visualizing the Electric Field

In this simulation, you can switch between presenting the electric potential or the EFLs

Notice how these two sets are perpendicular to each other at every point they meet?

 Electrostatics

      The Electric Field

            Electric Field Lines

Draw the EFLs for a dipole; use the equipotential surfaces as a guide

 Electrostatics

      The Electric Field

            Electric Field Lines

Read more about electric field lines, and the 5 criteria for a correct drawing.

Electrostatics

The Electric Field

Due to charge distributions

 Electrostatics

      The Electric Field

            Due to a straight-line-segment of length L

Segment x z Ex Ez
1
2
...
x
z

 Electrostatics

      The Electric Field

            Due to a straight-line-segment of length L

Detailed argument in Example 5.5

 Electrostatics

      The Electric Field

            Due to a straight-line-segment of length L

d\vec{E}_1=\frac{k\ dq}{r^2}\hat{r}_1
=\frac{k\ \lambda dl}{r^2}
(\cos\theta\hat{k}+\sin\theta\hat{i})
d\vec{E}_2=\frac{k\ dq}{r^2}\hat{r}_2
=\frac{k\ \lambda dl}{r^2}
(\cos\theta\hat{k}-\sin\theta\hat{i})
d\vec{E}_1+d\vec{E}_2=
\frac{k\ \lambda dl}{r^2} (2\cos\theta\hat{k})

 Electrostatics

      The Electric Field

            Due to a straight-line-segment of length L

d\vec{E}_1+d\vec{E}_2=
\frac{k\ \lambda dl}{r^2} (2\cos\theta\hat{k})
d\vec{E}=
\vec{E}=\int d\vec{E}= ?
=\int_x \frac{k\ \lambda dl}{r^2} (2\cos\theta\hat{k})
= \int_x \frac{ k\ \lambda\ dx}{(x^2+z^2)}\frac{2z}{\sqrt{x^2+z^2}} \hat{k}
=2k\lambda z \hat{k} \int_0^{L/2} \frac{dx}{(x^2+z^2)^{3/2}}
=\frac{k \lambda L}{z\sqrt{z^2+(L/2)^2}}\hat{k}

 Electrostatics

      The Electric Field

            Due to an infinite straight-line-segment

\vec{E}=\frac{k \lambda L}{z\sqrt{z^2+(L/2)^2}}\hat{k}

"Infinite" line-segment

L\gt\gt\gt z
\vec{E}\approx\frac{k \lambda L}{z\sqrt{(L/2)^2}}\hat{k}
L\gt\gt\gt z
z^2+(L/2)^2\approx (L/2)^2
=\frac{2k \lambda }{z}\hat{k}

 Electrostatics

      The Electric Field

            Due to a straight-line-segment of length L

\vec{E}=\frac{k \lambda L}{z\sqrt{z^2+(L/2)^2}}\hat{k}

"Infinite"ly far from the line-segment 

\vec{E}\approx\frac{k \lambda L}{z\sqrt{z^2}}\hat{k}
z\gt\gt\gt L
z^2+(L/2)^2\approx z^2
\approx\frac{k \lambda L}{z^2}\hat{k}
\approx\frac{k \ q_\text{total}}{z^2}\hat{k}

like a point-charge!

 Electrostatics

      The Electric Field

            Due to a line-segment in the shape of a ring

Detailed argument in Example 5.7

 Electrostatics

      The Electric Field

            Due to a ring of charge -- key points

=\lambda R\ d\theta
\vec{E}=\frac{k q_\text{total}z}{{(z^2+R^2)^{3/2}}}\hat{k}

Symmetry leads to Field only in z direction

What is the Electric Field at the center of the ring?

?

"Infinite"ly far from the ring? 

?
dq=\lambda \ dl

 Electrostatics

      The Electric Field

            Due to a disk of charge -- key points

\vec{E}=\int_{r'=0}^{r'=R} \frac{k\ q_\text{ring of radius r'}\ z}{\sqrt{(z^2+r'^2)^{3/2}}}\hat{k}

Symmetry leads to Field only in z direction

q_\text{ ring of radius r'}= \sigma \ 2\pi r\ dr

Electrostatics

The Electric Field

& The Electric Flux

 Electrostatics

      The Electric Field

            Definition of the Electric Flux

More details in Section 6.1

The Electric Flux through a given surface, s, is a measure of how much electric field passes through the surface.

 Electrostatics

      The Electric Field

            Definition of the Electric Flux

The Electric Flux through a given surface, s, is a measure of how much electric field passes through the surface.

For a flat surface, and a uniform field

\Phi=E_\perp \times dA
=\vec{E}\cdot\hat{n} \times dA
\text{the vector}
\text{Electric Field}
\text{the normal}
\text{to the surface}
\text{the component of}
\text{the Electric Field}
\text{normal to the surface}

 Electrostatics

      The Electric Field

            Definition of the Electric Flux

The Electric Flux through a given surface, s, is a measure of how much electric field passes through the surface.

In general, the field maybe variable, and the surface maybe curved.

so, surface is divided into patches, and the flux through the patches is added up

\Phi = \int_s \vec{E} \cdot \hat{n}\ d{A}
E_\perp \times dA
\vec{E}\cdot\hat{n} \times dA
\text{the normal}
\text{to the surface}
\text{the vector}
\text{Electric Field}
\Phi = \int_s \vec{E} \cdot d\vec{A}

 Electrostatics

      The Electric Field

            Definition of the Electric Flux

\Phi_1 = \int_{s_1} \vec{E} \cdot d\vec{A}
= \int_{s_1} E \hat{j} \cdot d{A}\hat{j}
= \int_{s_1} E\ dA
\Phi_2 = \int_{s_2} \vec{E} \cdot d\vec{A}
= \int_{s_2} E \hat{j} \cdot d{A}(\cos\theta\hat{j}+\sin\theta\hat{k})
= \int_{s_2} E\cos\theta\ dA
\Phi = \int_s \vec{E} \cdot d\vec{A}
= E\cos\theta A_2
= E\ A_1

Find the Electric Flux through the two surfaces shown in the figure due to a uniform field pointing in the +y direction.

h/\cos\theta
w
h

Which is larger?

\Phi_1 =EA_1=E\ wh
\Phi_2 =E \cos\theta A_2=(E\cos\theta)\ (wh/\cos\theta)=E\ wh=\Phi_1

Field is uniform:        is constant (magnitude and direction.)

Surface is flat:       is constant (direction.) 

\vec{E}
\hat{n}

 Electrostatics

      The Electric Field

            Definition of the Electric Flux

Field is radial w/ constant magnitude on the spherical surface.

Surface is spherical, so the normal to the surface is radial.

\Phi_s= \int_{s} \vec{E} \cdot d\vec{A}
= \int_{s} E \hat{r} \cdot d{A}\ \hat{r}
= \int_{s} E\ dA
\Phi = \int_s \vec{E} \cdot d\vec{A}
= E\ A_s

Find the electric flux through a spherical surface of radius R, centered around a charge +q.

= \frac{kq}{R^2}\cdot 4\pi R^2
= \frac{q}{\epsilon}

The Flux is independent of the size of the sphere!

 Electrostatics

      The Electric Field

            Gauss' Law

More details in Section 6.2

\Phi_\text{closed surface} = \frac{q_\text{enclosed}}{\epsilon}

 Electrostatics

      The Electric Field

            Gauss' Law -- example: uniformly charged sphere

More details in Example 6.6

 Electrostatics

      The Electric Field

            Gauss' Law -- example: conducting sphere

More details in Example 6.11

Electrostatics

The Electric Field

Wait, wait ... what is a field?

 Electrostatics

The topology of some geographical area can be represented by a field. In this case, the elevation above sea level is a scalar field. That is to say, for every point on this map, the physical quantity called the elevation above sea level has a known magnitude (represented by the contour lines.)

      The Electric Field

           What is a field?

 Electrostatics

The temperature distribution inside a coffee pot is a scalar field. i.e. at every point in the space within the pot, the physical quantity known as temperature has a given magnitude. (represented by the colormap)

      The Electric Field

           What is a field?

 Electrostatics

The gulf stream can be represented by a vector field. At every point on the ocean surface, the velocity of the water has some magnitude (represented by the color scale) and a direction (represented by the direction of the little arrows.

      The Electric Field

           What is a field?

 Electrostatics

A simulation of the electric field strength induced in a model of a human brain via external electrodes. The magnitude of the electric field at each location within the brain is represented by the color map. The direction of the electric field is not represented in this figure.

      The Electric Field

           What is a field?