Basic definition

An electric field is a region around a charged particle where other charged particles experience a force. Imagine you have a charged balloon (static electricity). If you bring another small charged object (like a tiny piece of paper) near the balloon, it either gets attracted to or repelled by the balloon. This interaction happens because of the electric field created by the balloon.

Visualization

Think of an electric field like the wind around a fan. When the fan is on, it creates a region where you can feel the wind. Similarly, a charged object creates an electric field around it. You can’t see the field, but you can feel its effects.

Here's what is looks like more precisely:

The magnetic field lines here go out of the positive charge and into the negative charge.

However, for charges of the same sign, the magnetic field lines steer away from each other:

Electric Field Lines

Electric fields are often represented by field lines. These lines show the direction a positive test charge would move if placed in the field. You can draw electric field lines by connecting the vectors shown above.

Here are some key points about electric field lines:

• They start on positive charges and end on negative charges.

• They never cross each other.

• The density of the lines indicates the strength of the electric field (closer lines mean a stronger field).

• Like charges repel each other, while opposite charges attract each other.

Electric Force and Electric Field Strength

The electric force is a force caused from a particle being in an electric field. If $q > 0$, the force goes in the same direction as the electric field. If $q < 0$ then the force goes in the opposite direction as the electric field.

$\vec {F}_e = \vec {E}q \quad [N]$

The electric field strength at a point in space is defined as the force ($\vec {F}_e$) experienced by a small positive test charge ($q$) placed at that point, divided by the magnitude of the charge.

$| \vec {E} | = \frac { | \vec {F}_e | } { q } \quad [ \frac {N} {C} \ \text {or} \ \frac {V} {m} ]$

The unit of electric field strength is newtons per coulomb ($\frac {N} {C}$) or volts per meter ($\frac {V} {m}$).

Example

Calculate the magnitude of the force applied onto an electron with an elemental charge of $e = 1.60 \times 10^{-19} C$ in the field of strength $1.25 \times 10^{8} \frac {N} {C}$.

Answer

$| \vec {F}_e | = | \vec {E} | q = (1.25 \times 10^{8} \frac {N} {C})(1.60 \times 10^{-19} C) = 2 \times 10^{-11} N$

Coulomb’s Law

To understand electric fields, we need to know Coulomb’s Law, which describes the force between two point charges:

$\vec {F}_e = k \frac { q_1 q_2 } { r^2 }$

Where $\vec {F}_e$ is the force between the charges, $k = 8.99 \times 10^9 \frac { N m^2 } { C^2 }$ is Coulomb's Constant, $q_1$ and $q_2$ are the magnitudes of both charges and $r$ is the distance between them.

From the equation above, we can also derive another equation for the strength of the electric field:

$| \vec {E} | = k \frac {q} { r^2 }$

Example

For two particles separated by distance of $5.00m$ with an equal magnitude of charge of $6.42 \times 10^{-6} C$:

a) What is the magnitude of the electric force applied onto them?

b) If the particles are both positively charged, is the force repelling or attracting the particles?

c) If the particles are both negatively charged, is the force repelling or attracting the particles?

d) If one particle is positively charged and the other is negatively charged, is the force repelling or attracting the particles?

Answer

a) We can use the equation we just used to calculate the magnitude of the electric force:

$\vec {F}_e = k \frac { q_1 q_2 } { r^2 } = (8.99 \times 10^9 \frac { N m^2 } { C^2 }) \frac { (6.42 \times 10^{-6} C)(6.42 \times 10^{-6} C) } { (5.00m)^2 } = 0.0148 N$

b) Since both particles are positively charged, they are like, meaning they repel each other.

c) Since both particles are negatively charged, they are like, meaning they repel each other.

d) Since one particle is positively charged and the other is negatively charged, they are unlike, meaning they attract each other.

Electric potential

Electric potential is a scalar quantity that represents the potential energy per unit charge at a specific point in an electric field. It gives a measure of how much work is needed to move a positive test charge from a reference point to a point in an electric field without producing any acceleration.

Here's how you calculate it:

$V = \frac { E_p } { q } \quad [V]$

Where $V$ is the electric potential, $E_p$ is the electric potential energy, and $q$ is the charge. It is in the units of Volts $(V)$

Electric potential difference

The electric potential difference is the difference of the electric potentials at two different points in an electric field.

The equation below describes the difference in electric potential energy for points $A$ and $B$:

$V_{AB} = V_B - V_A$

Just like how $V_{AB}$ is relative to $V_A$ and $V_B$, both $V_A$ and $V_B$ are relative to some reference point as well. In most cases, that reference point will be somewhere infinitely far.

This means that:

$V_A = V_{ \infty A} = V_A - V_{ \infty } = V_A - 0 = V_A$

$V_B = V_{ \infty B} = V_B - V_{ \infty } = V_B - 0 = V_B$

Here's an image showing the electric potential at the charge compared to the electric potential at a point infinitely far from it:

As you can see, the electric potential difference $\Delta V$ starts at infinity, and ends at the point. This can help determine the total amount of work needed to get a positive test particle from infinity, to its center.

Electric potential energy

The electric potential energy can be derive by combining the equation for work, as well as the equation for the force of an electric field, we get:

$E_p = k \frac { q_1 q_2 } { r }$

Where $k = 8.99 \times 10^9 \frac { N m^2 } { C^2 }$ is Coulomb's Constant, $q_1$ is the source charge, $q_2$ is the test charge in the electric field, and $r$ is the distance between them.

By using this equation, we can derive another equation for electric potential:

$V = \frac { E_p } { q } = k \frac { q_1 q_2 } { r } \frac {1} {q_2} = k \frac { q_1 } { r }$

Example

Three charges and a point are placed in the form of a square with side length $1.00m$ as shown below. The charges of the particles are as follows: $q_1 = 2.00 \mu C$, $q_2 = 4.00 \mu C$, $q_3 = -2.00 \mu C$. Calculate the electric potential at point P.

Answer

To find the potential at point $P$, we can calculate the individual potentials that occur due to each charge individually, then add them up:

$V_{ \text {total} } = V_1 + V_2 + V_3$

$V_1 = k \frac {q_1} {r} = (8.99 \times 10^9 \frac { N m^2 } { C^2 }) \frac { (2.00 \times 10^{-6} C) } { (1.00m) } = 1.80 \times 10^4 V$

$V_2 = k \frac {q_2} {r} = (8.99 \times 10^9 \frac { N m^2 } { C^2 }) \frac { (4.00 \times 10^{-6} C) } { \sqrt {(1.00m)^2 + (1.00m)^2 } } = 2.54 \times 10^4 V$

$V_3 = k \frac {q_2} {r} = (8.99 \times 10^9 \frac { N m^2 } { C^2 }) \frac { (-2.00 \times 10^{-6} C) } { (1.00m) } = -1.80 \times 10^4 V$

$V_{ \text {total} } = V_1 + V_2 + V_3 = (1.80 \times 10^4 V) + (2.54 \times 10^4 V) + (-1.80 \times 10^4 V) = 2.54 \times 10^4 V$

Directions when going in/out the page

In order to represent a 3D figure on a 2D image, we use a convention for when a vector points into the page, or out of the page.

When a vector points into the page, it is denoted by an "x". For a vector pointing out of the page, it is denoted by a dot.