Physics · Chapter 20
Study notes aligned to the official NEB syllabus.
Every electric charge sets up a condition in the space around it. Any other charge brought into that space feels a force of attraction or repulsion without the two charges touching. The region in which this influence acts is called the electric field of the charge. This chapter defines the electric field and its strength, works out the field of a point charge, describes field lines, studies the motion of a charged particle in a uniform field, and develops the idea of electric flux and Gauss's law, applying the law to a charged sphere, a line charge, and a charged plane conductor.
Throughout, the constant of electrostatics is written using the permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$, so that:
$$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\ \text{N m}^2\text{C}^{-2}$$
An electric field is the region of space surrounding an electric charge within which another electric charge experiences an electric force. If a small test charge placed at a point feels a force, an electric field exists at that point; if it feels no force, there is no field there.
The field is a property of the space set up by the source charge, and it exists whether or not a second charge is present to detect it. We map and measure the field with a small positive test charge $q_0$, chosen small enough that it does not disturb the source charge that produces the field.
The electric field strength (also called electric field intensity) at a point is defined as the force experienced by a unit positive charge placed at that point.
If $\vec{F}$ is the force experienced by a small stationary positive test charge $q_0$ placed at a point in the field, the electric field strength $\vec{E}$ at that point is:
$$\vec{E} = \frac{\vec{F}}{q_0}$$
Nature and unit. Electric field strength is a vector quantity. Its direction is the direction of the force on a positive charge. Its SI unit is the newton per coulomb $(\text{N C}^{-1})$, which is equivalent to the volt per metre $(\text{V m}^{-1})$.
The dimensional formula of electric field strength is:
$$ \begin{aligned} [E] &= \frac{[\text{force}]}{[\text{charge}]} \ &= \frac{[\mathrm{M,L,T^{-2}}]}{[\mathrm{A,T}]} \ &= [\mathrm{M,L,T^{-3},A^{-1}}] \end{aligned} $$
A uniform electric field is one that has the same magnitude and the same direction at every point. It is represented by equally spaced, parallel straight field lines.
Rearranging the definition $\vec{E} = \vec{F}/q_0$, the force on any charge $q$ placed in a field $\vec{E}$ is:
$$\vec{F} = q\vec{E}$$
Direction of the force:
The magnitude of the force is $F = qE$, and in a uniform field of known strength this force is the same everywhere in the region, so a charge released in it moves with constant acceleration.
Problem: A charge of $5\ \mu\text{C}$ is placed in a uniform electric field of strength $2 \times 10^{4}\ \text{N C}^{-1}$. Find the force on the charge.
Given:
Formula: $F = qE$
Substitution:
$$ \begin{aligned} F &= (5 \times 10^{-6}) \times (2 \times 10^{4}) \ &= 10 \times 10^{-2} \ &= 0.1\ \text{N} \end{aligned} $$
Answer: The force on the charge is $\mathbf{0.1\ N}$, directed along the field (the charge is positive).
Consider a point charge $+q$ placed at a point $O$ in free space. Let $P$ be a point at distance $r$ from $O$, so that $OP = r$. To find the field at $P$, place a small positive test charge $q_0$ there. By Coulomb's law, the force on the test charge is:
$$F = \frac{1}{4\pi\varepsilon_0},\frac{q,q_0}{r^{2}}$$
By the definition of field strength, $E = F/q_0$, so the magnitude of the electric field at $P$ due to the point charge $q$ is:
$$\boxed{E = \frac{1}{4\pi\varepsilon_0},\frac{q}{r^{2}}}$$
The field points radially outward from a positive charge and radially inward towards a negative charge. In a medium of relative permittivity (dielectric constant) $K$, the field is reduced to:
$$E = \frac{1}{4\pi\varepsilon_0 K},\frac{q}{r^{2}}$$
Every electric charge sets up a condition in the space around it. Any other charge brought into that space feels a force of attraction or repulsion without the two charges touching. The region in which this influence acts is called the electric field of the charge. This chapter defines the electric field and its strength, works out the field of a point charge, describes field lines, studies the motion of a charged particle in a uniform field, and develops the idea of electric flux and Gauss's law, applying the law to a charged sphere, a line charge, and a charged plane conductor.
Throughout, the constant of electrostatics is written using the permittivity of free space , so that:
An electric field is the region of space surrounding an electric charge within which another electric charge experiences an electric force. If a small test charge placed at a point feels a force, an electric field exists at that point; if it feels no force, there is no field there.
The field is a property of the space set up by the source charge, and it exists whether or not a second charge is present to detect it. We map and measure the field with a small positive test charge , chosen small enough that it does not disturb the source charge that produces the field.
The electric field strength (also called electric field intensity) at a point is defined as the force experienced by a unit positive charge placed at that point.
If is the force experienced by a small stationary positive test charge placed at a point in the field, the electric field strength at that point is:
Nature and unit. Electric field strength is a vector quantity. Its direction is the direction of the force on a positive charge. Its SI unit is the newton per coulomb , which is equivalent to the volt per metre .
The dimensional formula of electric field strength is:
A uniform electric field is one that has the same magnitude and the same direction at every point. It is represented by equally spaced, parallel straight field lines.
Rearranging the definition , the force on any charge placed in a field is:
Direction of the force:
The magnitude of the force is , and in a uniform field of known strength this force is the same everywhere in the region, so a charge released in it moves with constant acceleration.
Problem: A charge of is placed in a uniform electric field of strength . Find the force on the charge.
Given:
Formula:
Substitution:
Answer: The force on the charge is , directed along the field (the charge is positive).
Consider a point charge placed at a point in free space. Let be a point at distance from , so that . To find the field at , place a small positive test charge there. By Coulomb's law, the force on the test charge is:
By the definition of field strength, , so the magnitude of the electric field at due to the point charge is:
The field points radially outward from a positive charge and radially inward towards a negative charge. In a medium of relative permittivity (dielectric constant) , the field is reduced to: