Charged Particle in Uniform Electric Field | StudySmarter

As you may well know, an electric field around a point charge follows an inverse square law meaning that its strength decreases quadratically with distance. But did you know, through some clever geometry, it’s possible to construct an electric field with the same strength wherever you are? Such fields are known as uniform electric fields and are essential for capacitor technologies. In this article, we’ll look at uniform electric fields and their effect on charged particles moving through them. As we’ll see, the motion of charges within these fields is strikingly similar to the motion of objects under the earth’s gravitational pull.

Force on a Charged Particle in a Uniform Field

To understand the motion of a charged particle in a uniform field, we first need to look at how a uniform field can arise and the geometry of the field lines. As we have seen, a uniform field is one in which the field strength \(|E|\) does not vary from place to place. This means that the field lines of a uniform field should all be parallel and equally spaced, as seen in the figure below.

Fig. 1 – The field lines in a uniform electric field are parallel and evenly spaced.

A common example of such a field is between two oppositely charged parallel plates. To see how two parallel plates can produce a uniform electric field, consider the plates formed of a row of point charges. The electric field around a point charge \(q\) is given by the formula

\[|E|=\frac{1}{4\pi\epsilon_0}\frac{|q|}{r^2}\]

\(\epsilon_0\) is a physical constant known as the permittivity of free space, it has a value of \(\epsilon_0=8.85\times10^{-12}\,\mathrm{F}/\mathrm{m}\)

This means the electric field lines around a point charge point radially inwards or outwards depending on the charge of the particle. The overall electric field of a collection of point charges is then found by adding up all these radial field lines. In the case of a row of point charges within a plate, all the field components pointing parallel to the plate cancel out leaving a field that points perpendicular to the plate’s surface. This gives rise to the evenly spaced parallel field lines between the two plates. The strength of a uniform electric field between two plates with a uniform charge per unit area \(\sigma=\frac{Q}{A}\) is defined as

\[E=\frac{\sigma}{\epsilon_0}\]

Fig. 2 – The vertical components of the point charges electric fields all cancel out such that only parallel horizontal field lines are left.

This electric field \(E\) is produced because there is a potential difference \(V\) between the two plates, due to the build of opposite charges on each plate. This is expressed by the equation

\[E=\frac{V}{d}\]

where \(d\) is the distance between the two plates.

By convention, the potential difference is measured from the positive plate, increasing as you move towards the negative plate. Hence, the potential difference is parallel to the electric field, meaning that the potential isolines are surfaces perpendicular to the electric field lines and there is no component of the field along these isolines. As the electric field is uniform, these isolines are evenly spaced as the change in potential difference over distance remains constant.

Potential Isolines are surfaces along which the electric potential remains constant, similar to contour lines on a map where the altitude remains constant.

Fig. 3 – The potential isolines between two parallel plates are shown in purple, no work is done by the electric field when a charge is moved along these isolines hence they are perpendicular to the field lines

The force experienced by a particle with charge \(q\) as it moves a distance \(\Delta r\) between these two plates is given by

\[F=q\frac{\Delta V}{\Delta r}\]

where \(\Delta V\) is the change in potential over a distance of \(\Delta r\). Note that the force on a charged particle in an electric field can also be given as\[\begin{align}F&=qE\\\Rightarrow E&=\frac{\Delta V}{\Delta r}\end{align}\]

As \(E\) is a constant in a uniform field, we see that the force on a charged particle within a uniform electric field is also a constant. For example, an electron moving between the plates experiences a constant force towards the positive plate. This force always acts perpendicularly to the surface of the plates, there is never any force along the potential isolines as if there is no change in potential there can be no force.

Calculate the force on an electron if it moves between two parallel plates with a potential difference of \(150\,\mathrm{V}\) separated by a distance of \(0.5\,\mathrm{m}\).

First, we need to find the strength of the electric field between the two plates.

\[E=\frac{V}{d}=\frac{150\,\mathrm{V}}{0.5\,\mathrm{m}}=300\,\mathrm{N}/\mathrm{C}\]

To find the force we simply multiply the electric field strength by the charge of an electron \(q_{\mathrm{e}}=-1.6\times10^{-19}\,\mathrm{C}\).

\[F=q_eE=-1.6\times10^{-19}\,\mathrm{C}\cdot300\,\mathrm{N}/\mathrm{C}=-4.8\times10^{-17}\,\mathrm{N}\]

Acceleration of a Charged Particle in a Uniform Field

With the force on a charged particle in a uniform field established, the acceleration of a particle can be easily established from Newton’s second law.

\[\begin{align}F=&ma\\\Rightarrow \frac{qV}{d}=&ma\\\Rightarrow a=&\frac{qV}{dm}\end{align}\]

As the direction of acceleration is parallel to the force, we see that charged particles accelerate toward the plate with an opposite charge.

What is the acceleration of a proton moving in a uniform electric field produced by two parallel plates with a \(120\,\mathrm{V}\) potential difference separated by \(0.15\,\mathrm{m}\)?

Using the equation given above, and the mass and charge of a proton \[m_\mathrm{p}=1.67\times10^{-27}\,\mathrm{kg},\, q_\mathrm{p}=1.6\times10^{-19}\,\mathrm{C},\] we find the acceleration to be

\[a=\frac{1.6\times10^{-19}\,\mathrm{C}\cdot120\,\mathrm{V}}{0.15\,\mathrm{m}\cdot1.67\times10^{-27}\,\mathrm{kg}}=7.6\times10^{10}\,\mathrm{m}/\mathrm{s}^{2}\]

The Motion of a Charged Particle in a Uniform Field

Having found the force and acceleration of a charged particle in a uniform field, we can now look at how these forces affect the motion of a particle between two parallel plates. The motion of an initially stationary charged particle in a uniform field is particularly simple, the particle will be accelerated towards the plate of opposite charge in a straight line along a field line.

Fig. 4 – If a positive particle is initially stationary it will be accelerated by the uniform field, moving towards the negative plate along a field line.

Now, consider the case where a particle moves with a constant velocity between the plates, perpendicular to the direction of the electric field. This time there are two components of motion, one parallel to the surface of the plate due to the initial velocity (denoted \(x\)), and one perpendicular to the plate’s surface due to the force of the electric field (denoted \(y\)). We can use the kinematic equations to find the motion of the particle, with \(x,v_0,a\) denoting the displacement, initial velocity, and acceleration along each component,

\[\begin{align}x=&v_{x0}t+\frac{1}{2}a_xt^2\\y=&u_{y0}t+\frac{1}{2}a_yt^2\end{align}\]

If the particle has a constant initial velocity parallel to the plate surface, then \(a_x=0\) and \(u_y=0\). Using the equation for acceleration we found in the previous section, we find that

\[\begin{align}x=&v_{x0}t\\y=&\frac{1}{2}\frac{qV}{dm}t^2\end{align}\]

As \(v_{x0}\) and \(\frac{1}{2}\frac{qV}{dm}\) are both constants we find that \[y\propto x^2\] which demonstrates that a charged particle will follow a parabolic path through a uniform electric field. This is much like the case of throwing a ball horizontally, where the force of gravity pulls it towards the earth causing the ball to take a parabolic path. Such a path will always occur when an object experiences a constant force perpendicular to its constant velocity.

Fig. 5 – A positively charged particle moving at constant velocity through a uniform electric field will follow a parabolic path as shown in purple.

Kinetic Energy of a Charged Particle in a Uniform Field

The kinetic energy of a charged particle in a uniform field can be calculated in several ways. We could use kinematic equations and the expression for the acceleration of the particle in the field to find its velocity, from which the kinetic energy can be easily found. However, it is much simpler to find the kinetic energy of the particle by applying the Work-Energy Theorem.

The Work-Energy Theorem states that the net work done by the electric field on the particle is equal to the change in kinetic energy of the particle.

\[W_{\mathrm{net}}=\Delta K\]

The work done by a uniform electric field to move a particle with charge \(q\) from position \(\mathrm{A}\) to position \(\mathrm{B}\) can be found by multiplying the charge by the change in potential between these two positions.\[W_{\mathrm{AB}}=q\Delta V_{\mathrm{AB}}\]

This means that the change in kinetic energy of a particle in a uniform field as it moves from \(A\) to \(B\) is given by\[\Delta K_{\mathrm{AB}}=q\Delta V_{\mathrm{AB}}\]

Motion of a Charged Particle in a Uniform Field: Problems

Let’s take a look at some practice problems.

Q: Consider two parallel plates a distance of \(0.6\,\mathrm{m}\) apart. What would the voltage need to be between the two parallel plates in order for an electron to experience a force of \(500\,\mathrm{N}\)?

A: First re-arrange the equation for the force on a charged particle in a uniform field to find an expression for the voltage.\[\begin{align}F=&q_\mathrm{e}\frac{V}{d}\\\Rightarrow V=&\frac{Fd}{q_\mathrm{e}}\end{align}\] Plugging in the values from the question gives the voltage as\[V=\frac{500\,\mathrm{N}\cdot0.6\,\mathrm{m}}{-1.6\times10^{-19}\,\mathrm{C}}=-1.88\times10^{21}\,\mathrm{V}\]

Q: Two parallel plates a distance of \(0.3\,\mathrm{m}\) apart produce a uniform electric field with magnitude \(300\,\mathrm{N}\,\mathrm{C}^{-1}\). If a proton begins at rest at the positive plate and is accelerated towards the positive plate, what will its kinetic energy be just before it reaches the positive plate?

A: The work-kinetic energy theorem says that the work done by the electric field is equal to the negative of the change in kinetic energy of the particle. \[W_{\text{field}}=-\Delta K_{\text{particle}}\]So, first, we need to calculate the work done on the field to accelerate the particle as this will be equal to the final kinetic energy of the particle if it begins at rest. The work done by a uniform electric field, in moving the particle from one plate to another, is equivalent to the negative of the potential difference multiplied by the charge of the particle.\[W=-q_\mathrm{e}V\]If we choose the negative plate to have zero potential, then the potential difference will be positive as negative work is done to move the positive particle toward the negative plate.

We can find the potential difference between the two plates from the definition of a uniform electric field\[\begin{align}E&=\frac{V}{d}\\\Rightarrow V&=Ed\\&=300\,\mathrm{N}\,\mathrm{C}^{-1}\cdot0.3\,\mathrm{m}\\&=90\,\mathrm{V}\\\Rightarrow W&=-1.6\times10^{-19}\,\mathrm{C}\cdot 90\,\mathrm{V}\\&=-14.4\times10^{-18}\,\mathrm{J}\end{align}\]

Therefore, using the work-kinetic energy theorem, the final kinetic energy is\[\begin{align}\Delta K&=K_\mathrm{f}-K_\mathrm{i}\\&=K_\mathrm{f}=-W=14.4\times10^{-18}\,\mathrm{J}\end{align}\]

Q: Consider a proton moving directly in the middle of two parallel plates, parallel to their surfaces. The plates are \(1\mathrm{m}\) apart with a potential difference of \(0.01\,\mathrm{V}\) across them. If each plate is \(0.5\,\mathrm{m}\) long, how fast must the proton be moving initially to ensure it passes out the other side of the plates without first being pulled into contact with the negative plate?

A: To solve this problem, we need to split the particle’s motion into two components. One component \(x\) runs parallel to the plate surface, and the other \(y\) is perpendicular to the plate’s surface. We then find how long it will take for the particle to fall the \(0.5\,\mathrm{m}\) in the \(y\) direction. Then using this time, we can see how fast the article must move in the \(x\) direction so that it clears the \(0.5\,\mathrm{m}\) length of the plates.This requires the following kinematic equation\[x=v_0t+\frac{1}{2}at^2\]where \(x\) is the displacement of the particle, \(v_0\) is the initial velocity, and \(a\) is the acceleration of the particle.First applying the equation to the \(y\) direction, noting that there is no initial velocity in this direction.\[\begin{align}y&=\frac{1}{2}at^2\\\Rightarrow t&=\sqrt{\frac{2y}{a}}\end{align}\]We know the acceleration due to the electric field is given by\[\begin{align}a=&\frac{q_\mathrm{e}V}{dm}\\a=&\frac{1.6\times10^{-19}\,\mathrm{C}\cdot0.01\,\mathrm{V}}{1\,\mathrm{m}\cdot1.67\times10^{-27}\,\mathrm{kg}}\\=&0.95\times10^{6}\,\mathrm{m}\,\mathrm{s}^{-2}\end{align}\]So the time taken before the proton is pulled to the plate is\[\begin{align}t&=\sqrt{\frac{2y}{a}}=\sqrt{\frac{2\cdot0.5\,\mathrm{m}}{0.95\times10^{6}\,\mathrm{m}\,\mathrm{s}^{-2}}}\\&=\sqrt{0.95\times10^{-6}\,\mathrm{s}}=0.97\times10^{-3}\,\mathrm{s}\end{align}\]Now looking at the same kinematic equation in the \(x\) component, noting that there is no acceleration in this direction, we can plug in the time found above to give the minimum velocity required for the particle to clear the length of the plates.\[\begin{align}x&=v_{x0}t\\\Rightarrow v_0&=\frac{x}{t}\\&=\frac{1\,\mathrm{m}}{0.97\times10^{-3}\,\mathrm{s}}\\&=0.97\times10^{3}\,\mathrm{m}\,\mathrm{s}^{-1}\end{align}\]

To solve this problem, we need to split the particle’s motion into two components. One component \(x\) runs parallel to the plate surface, and the other \(y\) is perpendicular to the plate’s surface. We then find how long it will take for the particle to fall the \(0.5\,\mathrm{m}\) in the \(y\) direction. Then using this time, we can see how fast the article must move in the \(x\) direction so that it clears the \(0.5\,\mathrm{m}\) length of the plates.This requires the following kinematic equation\[x=v_0t+\frac{1}{2}at^2\]where \(x\) is the displacement of the particle, \(v_0\) is the initial velocity, and \(a\) is the acceleration of the particle.First applying the equation to the \(y\) direction, noting that there is no initial velocity in this direction.\[\begin{align}y&=\frac{1}{2}at^2\\\Rightarrow t&=\sqrt{\frac{2y}{a}}\end{align}\]We know the acceleration due to the electric field is given by\[\begin{align}a=&\frac{q_\mathrm{e}V}{dm}\\a=&\frac{1.6\times10^{-19}\,\mathrm{C}\cdot0.01\,\mathrm{V}}{1\,\mathrm{m}\cdot1.67\times10^{-27}\,\mathrm{kg}}\\=&0.95\times10^{6}\,\mathrm{m}\,\mathrm{s}^{-2}\end{align}\]So the time taken before the proton is pulled to the plate is\[\begin{align}t&=\sqrt{\frac{2y}{a}}=\sqrt{\frac{2\cdot0.5\,\mathrm{m}}{0.95\times10^{6}\,\mathrm{m}\,\mathrm{s}^{-2}}}\\&=\sqrt{0.95\times10^{-6}\,\mathrm{s}}=0.97\times10^{-3}\,\mathrm{s}\end{align}\]Now looking at the same kinematic equation in the \(x\) component, noting that there is no acceleration in this direction, we can plug in the time found above to give the minimum velocity required for the particle to clear the length of the plates.\[\begin{align}x&=v_{x0}t\\\Rightarrow v_0&=\frac{x}{t}\\&=\frac{1\,\mathrm{m}}{0.97\times10^{-3}\,\mathrm{s}}\\&=0.97\times10^{3}\,\mathrm{m}\,\mathrm{s}^{-1}\end{align}\]

Particle in a Uniform Electric Field – Key takeaways

• Uniform electric fields are electric fields where the field strength is the same everywhere within the field.
• Uniform electric fields arise between oppositely charged parallel plates, where their geometry ensures that the field lines are parallel and evenly spaced.
• The field strength of a uniform electric field is given by\[E=\frac{V}{d},\]where \(V\) is the potential difference across the parallel plates and \(d\) is the distance between them.
• We can use kinematic equations to analyze the motion of a charged particle in a uniform electric field, as it experiences constant acceleration.
• The change in kinetic energy of a charged particle as it moves between two points \(\mathrm{A}\) and \(\mathrm{B}\) is equal to the work done by the electric field on the particle given by\[\Delta K=W_{\mathrm{AB}}=q\Delta V_{\mathrm{AB}}\]

References

1. Fig. 1 – Uniform Electric Field, StudySmarter Originals
2. Fig. 2 – Uniform Field with Point Particles, StudySmarter Originals.
3. Fig. 3 Uniform Electric Field with Isolines, StudySmarter Originals.
4. Fig. 4 – Positive particle in Uniform Field, StudySmarter Originals.
5. Fig. 5 – Parabolic Motion of Charge in Uniform Field, StudySmarter Originals.