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Learning Introductory Physics with Activities

Paul J. Emigh, Rebecka Tumblin, Kathryn Hadley, Danielle Skinner

Section 21.4 Electric Field of Many Charges

The electric field is a vector field and vectors obey the principle of superposition. If multiple point charges exist in some region of space, and we want to know the electric field at some point in space in the vicinity of these charges, then we need to sum the vector components of the electric field from each charge at that point in space to determine the total electric field. It may be useful for you to review vector addition before moving forward.

Definition 21.4.1. Electric Field of Many Point Charges.

The electric field due to multiple point charges is the vector sum
\begin{equation*} \vec{E}_{tot} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + ... = \Sigma{\vec{E}_i} \text{.} \end{equation*}
where \(i \) refers to the electric field from the ith charge in the sum.
Figure 21.4.2. A visualization of the electric field at a point in space due to multiple charges.

Exercises Activities

1. Two Protons.

Consider two protons separated by a distance \(d \) on the \(x\)-axis equidistant from the origin.
  1. Overlaid on the appropriate coordinate system, sketch a diagram of the physical situation and label all quantities of interest.
  2. Considering the principle of superposition, if you are at a point on the positive \(y\)-axis, in which direction do you expect the net electric field to point? Back up your argument with a diagram.
  3. Use your coordinate system and defined variables from your diagram to construct a symbolic expression for the electric field \(\vec{E}(y)\) points along the \(y\)-axis of your coordinate system.
  4. What do you expect the field to look like if you move very far up the \(y\)-axis such that \(y \gg d\text{?}\)

2. Spreading out Charges.

A wire has four positive point charges located as shown in the figure below.
Figure 21.4.3. Four charges of charge \(+q\) are evenly distributed on a wire.
  1. What is the total charge on the wire?
  2. Suppose the wire were placed in a uniform electric field, \(\vec{E} = E_o \hat{y}\text{.}\) What is the net force on the wire?
  3. Imagine that the total charge on the wire was instead spread uniformly across the wire, from left to right. Does the net force change?
  4. Now imagine that the total charge was instead spread nonuniformly: much more charge on the right edge of the wire than the left edge. How does your strategy for finding the net force need to change?
Answer.
  1. The total charge is \(+4q\text{.}\) You can simply add up the charges!
  2. The electric force is \(\vec{F}^E = q \vec{E}\text{.}\) The total charge on the wire is \(+4q\text{,}\) so the force is \(\vec{F}^E = (+4q)E_o \hat{y}\text{.}\)
  3. The net force does not change. The same amount of charge is evenly distributed across the wire, just like before (but in smaller chunks of charge), so the force does not change.
  4. We must integrate the force!
    \begin{equation*} \vec{F} = q\vec{E} \end{equation*}
    \begin{equation*} d\vec{F} = \vec{E}dq \end{equation*}
    Using the linear charge density, we rewrite \(dq\text{:}\)
    \begin{equation*} d\vec{F} = \vec{E} \lambda(x) dx \end{equation*}
    Integrating both sides of this expression:
    \begin{equation*} \vec{F} = \int \vec{E} \lambda(x) dx \end{equation*}
    Since the electric field doesn’t change along x, it can be pulled out of the integral:
    \begin{equation*} \vec{F} = \vec{E} \int \lambda(x) dx \end{equation*}
    The integral of \(\lambda(x)dx\) gives you the total charge. It turns out, the electric force is the same, because the electric field is uniform and the total charge remained the same!
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