Title: Default for 9.3 The scalar product
Course: HELM Workbooks 2023-2024

Definition of the scalar product

Consider the two vectors \(\underline{a}\) and \(\underline{b}\) shown in the figure below.

[Image]
Two vectors subtend an angle \(\theta\)

Note that the tails of the two vectors coincide and that the angle between the vectors is labelled \(\theta\). Their scalar product, denoted by \(\underline{a}\cdot\underline{b}\), is defined as the product \(|\underline{a}|\,|\underline{b}|\,\cos\theta\). It is very important to use the dot in the formula. The dot is the specific symbol for the scalar product and is the reason why the scalar product is also known as the dot product. You should not use a \(\times\) sign in this context because this sign is reserved for the vector product which is quite different.

The angle \(\theta\) is always chosen to lie between 0 and \(\pi\), and the tails of the two vectors must coincide. The figure below shows two incorrect ways of measuring \(\theta\).

[Image: tails of vectors don't coincide and does not lie between 0 and pi]
\(\theta\) should not be measured in these ways

Key Point

\[\textbf{The scalar product} \hbox{ of ${\underline a}$ and ${\underline b}$ is: }\quad \underline{a}\cdot\underline{b} =|\underline{a}|\,|\underline{b}|\,\cos\theta\] We can remember this formula as:

“The modulus of the first vector, multiplied by the modulus of the second vector, multiplied by the cosine of the angle between them.”

Clearly \(\underline{b}\cdot\underline{a}=|\underline{b}|\,|\underline{a}|\,\cos\theta\) and so \[\qquad\underline{a}\cdot\underline{b}=\underline{b}\cdot\underline{a}.\] Thus we can evaluate a scalar product in any order: the operation is commutative.

Example

Vectors \(\underline{a}\) and \(\underline{b}\) are shown in the figure below. Vector \(\underline{a}\) has modulus 6 and vector \(\underline{b}\) has modulus 7 and the angle between them is \(60^\circ\). Calculate \(\underline{a} \cdot \underline{b}\).

[Image]

Solution

The angle between the two vectors is \(60^\circ\). Hence

\(\qquad \underline{a}\cdot\underline{b}= |\underline{a}|\,|\underline{b}|\,\cos\theta=(6)(7)\cos 60^\circ=21\)

The scalar product of \(\underline{a}\) and \(\underline{b}\) is 21. Note that a scalar product is always a scalar.

Example

Find \(\underline{i}\cdot\underline{i}\) where \(\underline{i}\) is the unit vector in the direction of the positive \(x\) axis.

Solution

Because \(\underline{i}\) is a unit vector its modulus is 1. Also, the angle between \(\underline{i}\) and itself is zero. Therefore

\(\qquad \underline{i}\cdot\underline{i} = (1)(1)\cos 0^\circ = 1\)

So the scalar product of \(\underline{i}\) with itself equals 1. It is easy to verify that \(\underline{j}\cdot\underline{j}=1\) and \(\underline{k}\cdot\underline{k}=1\).

Example

Find \(\underline{i}\cdot\underline{j}\) where \(\underline{i}\) and \(\underline{j}\) are unit vectors in the directions of the \(x\) and \(y\) axes.

Solution

Because \(\underline{i}\) and \(\underline{j}\) are unit vectors they each have a modulus of 1. The angle between the two vectors is \(90^\circ\). Therefore

\(\qquad \underline{i}\cdot\underline{j}=(1)(1)\cos 90^\circ = 0\)

That is \(\underline{i}\cdot\underline{j}=0\).

The following results are easily verified:

Key Point

\[\begin{aligned} \underline{i}\cdot\underline{i} = \underline{j}\cdot\underline{j} = \underline{k}\cdot\underline{k}=1\\ \underline{i}\cdot\underline{j} = \underline{j}\cdot\underline{i} = 0\\ \underline{i}\cdot\underline{k} = \underline{k}\cdot\underline{i} = 0\\ \underline{j}\cdot\underline{k} = \underline{k}\cdot\underline{j} = 0\end{aligned}\]

Generally, whenever any two vectors are perpendicular to each other their scalar product is zero because the angle between the vectors is \(90^\circ\) and \(\cos 90^\circ=0\).

Key Point

The scalar product of perpendicular vectors is zero.

Resolving one vector along another

The scalar product can be used to find the component of a vector in the direction of another vector. Consider the figure below which shows two arbitrary vectors \(\underline{a}\) and \(\underline{n}\). Let \(\underline{\hat{n}}\) be a unit vector in the direction of \(\underline{n}\).

[Image: projection of a onto n]

Study the figure carefully and note that a perpendicular has been drawn from \(P\) to meet \(\underline{n}\) at \(Q\). The distance \(OQ\) is called the projection of \(\underline{a}\) onto \(\underline{n}\). Simple trigonometry tells us that the length of the projection is \(|\underline{a}|\cos\theta\). Now by taking the scalar product of \(\underline{a}\) with the unit vector \(\underline{\hat{n}}\) we find

  \(\qquad\underline{a}\cdot\underline{\hat{n}} =|\underline{a}|\,|\hat{\underline{n}}|\,\cos\theta =|\underline{a}|\cos\theta\qquad\qquad\hbox{(since } |\underline{\hat{n}}|=1)\)

 We conclude that

Key Point

\[\underline{a}\cdot \underline{\hat{n}} \hbox{ is the component of } \underline{a} \hbox{ in the direction of } \underline{n}\]

Example

The figure above shows a plane containing the point \(A\) which has position vector \(\underline{a}\). The vector \(\underline{\hat{n}}\) is a unit vector perpendicular to the plane (such a vector is called a unit normal vector). Find an expression for the perpendicular distance, \(\ell\), of the plane from the origin.

[Image]

Solution

From the diagram we note that the perpendicular distance \(\ell\) of the plane from the origin is the projection of \(\underline{a}\) onto \(\underline{\hat{n}}\) and, using the previous Key point, is thus \(\underline{a}\cdot \hat{\underline{n}}\).

Vectors and electrostatics

Electricity is important in several branches of engineering – not only in electrical or electronic engineering. For example, the design of the electrostatic precipitator plates for cleaning the solid fuel power stations involves both mechanical engineering (structures and mechanical rapping systems for cleaning the plates) and electrostatics (to determine the electrical forces between solid particles and plates).

The following example and tasks relate to the electrostatic forces between particles. Electric charge is measured in coulombs (C). Charges can be either positive or negative.

The force between two charges

Let \(q_1\) and \(q_2\) be two charges in free space located at points \(P_1\) and \(P_2\). Then \(q_1\) will experience a force due to the presence of \(q_2\) and directed from \(P_2\) towards \(P_1\).

This force is of magnitude \(K \dfrac{q_1q_2}{r^2}\) where \(r\) is the distance between \(P_1\) and \(P_2\) and \(K\) is a constant.

In vector notation this coulomb force (measured in newtons) can then be expressed as \(\underline{F}=K\dfrac{q_1q_2}{r^2} \hat{\underline{ r}}\) where \(\hat{\underline{ r}}\) is a unit vector directed from \(P_2\) towards \(P_1\).

The constant \(K\) is known to be \(\dfrac{1}{4\pi\varepsilon_0}\) where \(\varepsilon_0=8.854\times 10^{-12}\) F m\(^{-1}\) (farads per metre).

The electric field

A unit charge located at a general point \(G\) will then experience a force \(\dfrac{Kq_1}{r_1^2}\hat{\underline{  r}}_1\) (where \(\hat{\underline{ r}}_1\) is the unit vector directed from \(P_1\) towards \(G\)) due to a charge \(q_1\) located at \(P_1\). This is the electric field \( \underline{E} \) newtons per coulomb (N C\(^{-1}\) or alternatively V m\(^{-1}\)) at \(G\) due to the presence of \(q_1\).

For several point charges \(q_1\) at \(P_1,\ q_2\) at \(P_2\) etc., the total electric field \( \underline{E} \) at \(G\) is given by

\(\underline{E}= \dfrac{K q_1}{r_1^2} \hat{\underline{ r}}_1+\dfrac{Kq_2}{r_2^2}\ \hat{\underline{ r}}_2+\ldots\)

where \(\hat{\underline { r}}_i\) is the unit vector directed from point \(P_i\) towards \(G\).

From the definition of a unit vector we see that

\(\underline{E}=\dfrac{Kq_1}{r_1^2} \dfrac{\underline{r_1}}{|\underline{r_1}|}+\dfrac{Kq_2}{r_2^2} \dfrac{\underline{r_2}}{|\underline{r_2}|}+\ldots=\dfrac{Kq_1}{|r_1|^3}\underline{r_1}+\dfrac{Kq_2}{|\underline{r_2}|^3}\underline{r_2}+\ldots\)

\(\underline{E}=\dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1}{|\underline{r_1}|^3}\underline{r_1}+\dfrac{q_2}{|\underline{r_2}|^3}\underline{r_2}+\ldots\right]\)

where \(\underline{r_i}\) is the vector directed from point \(P_i\) towards \(G\), so that \(\underline{r_1}=\underline{OG}-\underline{OP}_1\) etc., where \(\underline{OG}\) and \(\underline{OP}_1\) are the position vectors of \(G\) and \(P_1\) (see the figure below).

[Image]

\[\underline{OP_1}+\underline{P_1G}=\underline{OG}\qquad\qquad \underline{P_1G}=\underline{OG}-\underline{OP_1}\]

The work done

The work done \(W\) (energy expended) in moving a charge \(q\) through a distance \(\text{d} S\), in a direction given by the unit vector \({{\underline S}}/|{{\underline S}}|\), in an electric field \({\underline E}\) is (defined by)

\(\qquad W=-q{{\underline E}}.\text{d} {{\underline S}}\) where \(W\) is in joules.

A formula for finding the scalar product

We can use the results summarized previously to obtain a formula for finding a scalar product when the vectors are given in Cartesian form. We consider vectors in the \(xy\)-plane. Suppose \(\underline{a}=a_1\underline{i} + a_2\underline{j}\) and \(\underline{b}=b_1\underline{i}+b_2\underline{j}\). Then

\[\begin{aligned} \underline{a}\cdot\underline{b}&=(a_1\underline{i}+a_2\underline{j})\cdot(b_1\underline{i}+b_2\underline{j})\\ &= a_1\underline{i}\cdot(b_1\underline{i}+b_2\underline{j}) + a_2\underline{j}\cdot(b_1\underline{i}+b_2\underline{j})\\ &= a_1b_1\underline{i}\cdot\underline{i} + a_1b_2\underline{i}\cdot\underline{j} + a_2b_1\underline{j}\cdot\underline{i}+a_2b_2\underline{j}\cdot\underline{j}\end{aligned}\]

Thus to find the scalar product of two vectors their \(\underline{ i}\) components are multiplied together, their \(\underline{ j}\) components are multiplied together and the results are added.

Using the results explained in a previous Key Point, we can simplify this to give the following formula:

Key Point

\[\hbox{If} \quad \underline{ a}=a_1\underline{ i} + a_2\underline{ j}\quad\hbox{ and }\quad \underline{ b}=b_1\underline{ i}+b_2\underline{ j}\quad \hbox{  then }\] \[\underline{ a}\cdot\underline{b}= a_1b_1+a_2b_2\]

Thus to find the scalar product of two vectors their \(\underline{ i}\) components are multiplied together, their \(\underline{ j}\) components are multiplied together and the results are added.

Example

If \(\underline{ a}=7\underline{ i}+8\underline{ j}\) and \(\underline{ b}=5\underline{ i}-2\underline{ j}\), find the scalar product \(\underline{a}\cdot\underline{b}\).

Solution

We use the previous Key Point: \(\qquad\underline{ a}\cdot\underline{ b} = (7\underline{ i}+8\underline{ j})\cdot(5\underline{i}-2\underline{ j})= (7)(5) + (8)(-2)= 35-16= 19\)

The formula readily generalises to vectors in three dimensions as follows:

Key Point

\[\hbox{If} \quad \underline{ a}=a_1\underline{ i} + a_2\underline{ j} + a_3\underline{ k}\quad \hbox{ and }\quad\underline{ b}=b_1\underline{ i}+b_2\underline{ j} + b_3\underline{ k} \hbox{  then }\] \[\underline{ a}\cdot\underline{b} = a_1b_1 + a_2b_2 + a_3b_3\]

Example

If \(\underline{ a}=5\underline{ i}+3\underline{ j}-2\underline{ k}\) and \(\underline{ b}=8\underline{i}-9\underline{ j}+11\underline{ k}\), find \(\underline{a}\cdot\underline{b}\).

Solution

We use the formula in the previous Key Point:

\(\qquad \underline{ a}\cdot\underline{b}=(5)(8) + (3)(-9) + (-2)(11) = 40-27-22=-9\)

Note again that the result is a scalar: there are no \(\underline{ i}\)’s, \(\underline{ j}\)’s, or \(\underline{k}\)’s in the answer.

The above result is generally true:

Key Point

 For any vector \(\underline{ r}\),     \(|\underline{r}|^2=\underline{ r}\cdot\underline{ r}\)

Using the scalar product to find the angle between vectors

We have two distinct ways of calculating the scalar product of two vectors. As we hope you will remember \(\underline{a}\cdot\underline{b} =|\underline{a}|\,|\underline{b}|\,\cos\theta\) whilst from a previous key point \(\underline{ a}\cdot\underline{b} = a_1b_1 + a_2b_2 + a_3b_3\). Both methods of calculating the scalar product are entirely equivalent and will always give the same value for the scalar product. We can exploit this correspondence to find the angle between two vectors. The following example illustrates the procedure to be followed.

Example

Find the angle between the vectors \(\underline{ a}=5\underline{ i}+3\underline{ j}-2\underline{ k}\) and \(\underline{ b}=8\underline{ i}-9\underline{ j}+11\underline{ k}\).

Solution

The scalar product of these two vectors is \(-9\). The modulus of \(\underline{ a}\) is \(\sqrt{5^2+3^2+(-2)^2}=\sqrt{38}\). The modulus of \(\underline{ b}\) is \(\sqrt{8^2+(-9)^2+11^2}=\sqrt{266}\). Substituting these values for \({\underline a}\cdot {\underline b},\ |{\underline a}|\) and \(|{\underline b}|\) into the formula for the scalar product we find

\[\begin{aligned} \underline{ a}\cdot\underline{b} &= |\underline{ a}|\,|\underline{ b}|\,\cos \theta \\ -9 &= \sqrt{38}\,\sqrt{266}\cos\theta\end{aligned}\] from which \[\displaystyle{\quad\cos\theta = {-9\over \sqrt{38}\sqrt{266}} = -0.0895}\] so that \[\quad\theta = \cos^{-1}(-0.0895)=95.14^\circ.\]

In general, the angle between two vectors can be found from the following formula:

Key Point

The angle \(\theta\) between vectors \(\underline{a}\), \(\underline{b}\) is such that:

\[\cos\theta = {\underline{ a}\cdot\underline{b}\over |\underline{ a}|\,|\underline{ b}|}\]

Exercise

If \(\underline{ p}=[value]\underline{ i}-[value]\underline{ j}+7\underline{ k}\) and \(\underline{ q}=6\underline{ i}-\underline{ j}+[value]\underline{ k}\), then \(\underline{ p}\cdot\underline{ q}\) is:
Feedback: We use the formula given in a previous Key point: \(\underline{p}\cdot\underline{q}=([value])(6) + (-[value])(-1) + (7)([value]) = [value]+[value]+[value]=[value]\) Note again that the result is a scalar: there are no \(\underline{ i}\)'s, \(\underline{ j}\)'s, or \(\underline{ k}\)'s in the answer.

Exercise

If \(\underline{ r}=[value]\underline{ i}+[value]\underline{ j}+[value]\underline{ k}\), find \(\underline{ r}\cdot\underline{ r}\).

Now compute \(|\underline{ r}|^2\).
Feedback: We use the formula given in a previous Key point: \({\underline r}\cdot {\underline r}=([value]{\underline i}+[value]{\underline j}+[value]{\underline k})\cdot([value]{\underline i}+[value]{\underline j}+[value]{\underline k})=([value])([value])+([value])([value])+([value])([value])=[value]\). Then: \(|\underline{r}|=\sqrt{[value]+[value]+[value]}=\sqrt{[value]}\), hence \(|\underline{r}|^2= [value]\). Thus, illustrating that: \(|\underline{r}|^2=\underline{r}\cdot\underline{r}\).

Exercise

If \(\underline{a}=[value]\underline{i}-[value]\underline{j}\) and \(\underline{b}=3\underline{i}+[value]\underline{j}\) find \(\underline{a}\cdot\underline{b}\)

Now find \(\underline{b}\cdot\underline{a}\)
Feedback: We use the formula given previously: \(\underline{a}\cdot\underline{b}=([value])(3) + (-[value])([value]) = [value]-[value]=[value]\) And also remember that \(\underline{a}\cdot\underline{b} = \underline{b}\cdot\underline{a}\) so both answers are the same.

Exercise

Find the angle between \(\underline{p}=[value]\underline{i}+\underline{j}\) and \(\underline{q}=[value]\underline{i}+[value]\underline{j}\).

(Express your answer in degrees and round your answer up to one decimal place)

\(^\circ\) (degrees)
Feedback: First, we compute the modulus of \(\underline{ p}\) and \(\underline{ q}\): \[|\underline{p}| = \sqrt{[value]^2+1^2}=\sqrt{[value]}, \quad (|\underline{q}| = \sqrt{[value]^2+[value]^2}=\sqrt{[value]}.\] Then, we compute \( \underline{p}\cdot \underline{q} = ([value])(1)+([value])([value])=[value]\). Substituting these values for \({\underline p}\cdot {\underline q},|\underline{p}|\) and \(|\underline{q}|\) into the formula for the scalar product we have: \[\underline{p}\cdot\underline{q}=|\underline{p}|\,|\underline{q}| \cos\theta \Rightarrow [value] = \sqrt{[value]} \sqrt{[value]} \cos \theta \] which is equivalent to \( \cos \theta = \dfrac{[value]}{\sqrt{[value]}}. \) Finally we find that that: \[ \theta = \cos^{-1} \left( \dfrac{[value]}{\sqrt{[value]}} \right) = [value]^\circ.\]

Exercise

If \(\underline{a}\) and \(\underline{b}\) are perpendicular, simplify \((\underline{a}-[value]\underline{b})\cdot(3\underline{a}+[value]\underline{b})\).

Please write your answer as an algebraic expression in terms of \(a^2\) and \(b^2\) recalling that \(\underline{a} \cdot \underline{a} = a^2\) and \(\underline{b} \cdot \underline{b} = b^2\). Write the modulus of \(\underline{a}\) and \(\underline{b}\) as a and b, respectively.
Feedback: We proceed to compute \((\underline{a}-[value]\underline{b})\cdot(3\underline{a}+[value]\underline{b})\) as follows: \( (\underline{a}-[value]\underline{b})\cdot(3\underline{a}+[value]\underline{b}) = 3 \underline{a} \cdot \underline{a} + [value] \underline{a} \cdot \underline{b} +(3)(-[value]) \underline{a} \cdot \underline{b} + (-[value])([value]) \underline{b} \cdot \underline{b} \) As \(\underline{a} \) and \(\underline{b}\) are perpendicular, their dot product is zero, thus: \( (\underline{a}-[value]\underline{b})\cdot(3\underline{a}+[value]\underline{b}) = 3 \underline{a} \cdot \underline{a} + (-[value])([value]) \underline{b} \cdot \underline{b} = 3a^2 - [value] b^2 \).

Exercise

The following vectors \([value]\underline{i}+[value]\underline{j}\) and \([value]\underline{i}-[value]\underline{j}\) perpendicular. True or False?
Feedback: If the scalar product of these two vectors is zero, then both vectors are perpendicular. Otherwise, they are not. In this case, \[([value]\underline{i} + [value]\underline{j}) \cdot ([value]\underline{i} + [value]\underline{j}) = [value] = [value].\] Therefore, the two vectors are [value].

Exercise

If \(\underline{ p}=\underline{ i}+[value]\underline{ j}+[value]\underline{ k}\) and \(\underline{ q}=3\underline{ i}-[value]\underline{ j}+[value]\underline{ k}\), find \(\underline{p}\cdot\underline{q}\).
Feedback: We use the formula given in a previous Key point: \(\underline{p}\cdot\underline{q}=(1)(3) + ([value])(-[value]) + ([value])([value]) = 3-[value]+[value]=[value]\) Note again that the result is a scalar: there are no \(\underline{ i}\)'s, \(\underline{ j}\)'s, or \(\underline{ k}\)'s in the answer.

Exercise

Find the angle between the vectors \(\underline{p}=[value]\underline{i}+\underline{j}+\underline{k}\) and \(\underline{q}=[value]\underline{i}+[value]\underline{j}+2\underline{k}\).

(Express your answer in degrees and round your answer up to one decimal place)

\(^\circ\) (degrees)
Feedback: First, we compute the modulus of \(\underline{ p}\) and \(\underline{ q}\): \(|\underline{p}| = \sqrt{[value]^2+1^2 +1^2}=\sqrt{[value]}\), \(|\underline{q}| = \sqrt{[value]^2+[value]^2 + 2^2}=\sqrt{[value]}\). Then, we compute \( \underline{p}\cdot \underline{q} = ([value])([value])+(1)([value]) + (1)(2)=[value]\). Substituting these values for \({\underline p}\cdot {\underline q},|\underline{p}|\) and \(|\underline{q}|\) into the formula for the scalar product \( \underline{p}\cdot\underline{q}=|\underline{p}|\,|\underline{q}| \cos\theta \) we have: \( [value] = \sqrt{[value]} \sqrt{[value]} \cos \theta \) which is equivalent to \( \cos \theta = \dfrac{[value]}{\sqrt{[value]}} \) finally having that \( \theta = \cos^{-1} \left( \dfrac{[value]}{\sqrt{[value]}} \right) = [value]^\circ\)

Exercise

The work done by a force \(\underline{F}\) in moving a body through a displacement \(\underline{r}\) is given by \(\underline{F}\cdot\underline{r}\). Find the work done by the force \(\underline{F}=3\underline{i}+[value]\underline{k}\) if it causes a body to move from the point with coordinates \((1,1,2)\) to the point \(([value],3,5)\).

units
Feedback: First, we proceed to compute the displacement vector \(\underline{r}\) subtracting the final point \(([value],3,5)\) minus the initial point \((1,1,2)\): \(\underline{r} = ([value],3,5) - (1,1,2) = [value]\underline{i} + 2\underline{j} + 3\underline{k}\) Then, we proceed to compute the work done by the force \(\underline{F}=3\underline{i}+[value]\underline{k}\) using the formula given in the statement: \(W = \underline{F}\cdot\underline{r} = (3)([value])+(0)(2)+([value])(3) = [value] \) units.

Engineering Example

In free space, point charge \(q_1 = 10\ \mathrm{nC}\) (\(1\ \mathrm{nC} = 10^{-9}\ \mathrm{C}\), i.e. a nanocoulomb) is at \(P_1(0,-4,0)\) and charge \(q_2 = 20\ \mathrm{nC}\) is at \(P_2=(0,0,4)\).

[Note: Since the \(x\)-coordinate of both charges is zero, the problem is two-dimensional in the \(yz\)-plane as shown in the following figure.]

[Image]

1. Find the field at the origin \(\underline{E}_{1,2}\) due to \(q_1\) and \(q_2\).
2. Where should a third charge \(q_3=30\ \mathrm{nC}\) be placed in the \(yz\) plane so that the total field due to \(q_1,q_2,q_3\) is zero at the origin?

Solution

1. Total field at the origin \(\underline{E}_{1,2}\) = (field at origin due to charge at \(P_1\)) + (field at origin due to charge at \(P_2\)). Therefore,
\[\begin{aligned} {\underline E}_{1,2} &= \dfrac{10\times 10^{-9}}{4\pi \times 8.854\times 10^{-12}\times 4^2}{\underline j}+\dfrac{20\times 10^{-9}}{4\pi\times 8.854\times 10^{-12}\times4^2}({-\underline k)}\\ &=5.617{\underline j}-11.23{\underline k} \end{aligned}\]
(The negative sign in front of the second term results from the fact that the direction from \(P_2\) to \(O\) is in the \(-z\) direction.)
2. Suppose the third charge \(q_3=30\ \mathrm{nC}\) is placed at \(P_3(0,a,b)\). The field at the origin due to the third charge is
\[{\underline E}_3=\dfrac{30\times 10^{-9}}{4\pi\times 8.854\times 10^{-12}\times (a^2+b^2)}\times\dfrac{-(a\underline j +b\underline k)}{(a^2+b^2)^{1/2}}\]
where \(\dfrac{a\underline{j}+b\underline{k}}{(a^2+b^2)^{1/2}}\) is the unit vector in the direction from \(O\) to \(P_3\)
If the position of the third charge is such that the total field at the origin is zero, then \({\underline E}_3=-{\underline E}_{1,2}\). There are two unknowns (\(a\) and \(b\)). We can write down two equations by considering the \(\underline{j}\) and \(\underline{k}\) directions.
\[{\underline E}_3=-269.6\left[\dfrac{a}{(a^2+b^2)^{3/2}}\underline{j}+\dfrac{b}{(a^2+b^2)^{3/2}}\underline{k}\right]\quad{\underline E}_{1,2}=5.617\underline{j}-11.23\underline{k}\]
So
\(\qquad 5.617=269.6\times \dfrac{a}{(a^2+b^2)^{3/2}}\)    (1)
\(\qquad -11.23=269.6\times \dfrac{b}{(a^2+b^2)^{3/2}}\)    (2)
So
\(\qquad \dfrac{a}{(a^2+b^2)^{3/2}}=0.02083\)    (3)
\(\qquad \dfrac{b}{(a^2+b^2)^{3/2}}=-0.04165\)    (4)
Squaring and adding (3) and (4) gives \(\quad \dfrac{a^2+b^2}{(a^2+b^2)^3}=0.002169\)
So
\(\qquad (a^2+b^2)=21.47\)    (5)
Substituting back from (5) into (1) and (2) gives \(a=2.07\) and \(b=-4.14,\) to \(3\) s.f.

Exercise

Eight point charges of \([value]\ \mathrm{nC}\) each are located at the corners of a cube in free space which is \(1\ \mathrm{m}\) on each side (see the figure below). Calculate \(|{\underline E}|\) at

1. the centre of the cube: \(\mathrm{V}\,\mathrm{m}^{-1}\)
2. the centre of any face: \(\mathrm{V}\,\mathrm{m}^{-1}\)
3. the centre of any edge: \(\mathrm{V}\,\mathrm{m}^{-1}\)

Note: Please write your answers with 2 decimal places

[Image: sketch of cube]
Feedback: 1. You should know that the field at the centre of the cube is zero because of the symmetrical distribution of the charges. 2. Because of the symmetrical nature of the problem it does not matter which face is chosen in order to find the magnitude of the field at the centre of a face. Suppose the chosen face has corners located at \(P(1,1,1), T(1,1,0), R(0,1,0)\) and \(S(0,1,1)\) then the centre (\(C\)) of this face can be seen from the diagram to be located at \(C\left(\dfrac{1}{2}, 1,\dfrac{1}{2}\right)\). The electric field at \(C\) due to the charges at the corners \(P,T,R\) and \(S\) will then be zero since the field vectors due to equal charges located at opposite corners of the square \(PTRS\) cancel one another out. The field at \(C\) is then due to the equal charges located at the remaining four corners \((OABD)\) of the cube, and we note from the symmetry of the cube, that the distance of each of these corners from \(C\) will be the same. In particular the distance \[OC=\sqrt{\left(\dfrac{1}{2}\right)^2+1^2+\left(\dfrac{1}{2}\right)^2}=\sqrt{1.5}\ \text{m}.\] The electric field \(\underline{E}\) at \(C\) due to the remaining charges can then be found using \(\underline{E}=\dfrac{1}{4\pi\varepsilon_0}\sum_1^4\dfrac{q_i\cdot \underline{r}_i}{|\underline{r}_i|^3}\) where \(q_1\) to \(q_4\) are the equal charges (\([value] \times 10^{-9}\) coulombs) and \(\underline{r}_1\) to \(\underline{r}_4\) are the vectors directed from the four corners, where the charges are located, towards \(C\). In this case since \(q_1=[value] \times 10^{-9}\) coulombs and \(|\underline{r}_i|=\sqrt{1.5}\) for \(i=1\) to \(i=4\) we have \(\underline{E}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{[value] \times 10^{-9}}{(1.5)^{3/2}}\left[\underline{r}_1+\underline{r}_2+\underline{r}_3+\underline{r}_4\right],\) where \(\underline{r}_1=\underline{AC}=\left[\begin{array}{c}\dfrac{1}{2}\\ 1\\ \dfrac{1}{2}\end{array}\right]-\left[\begin{array}{c}0\\0\\1\end{array}\right],\ \underline{r}_2=\underline{BC} =\left[\begin{array}{c}\dfrac{1}{2}\\ 1\\ \dfrac{1}{2}\end{array}\right]-\left[\begin{array}{c}1\\0\\1\end{array}\right] \text{etc.}\) Thus \(\underline{E}=\dfrac{1}{4\pi \varepsilon_0} \dfrac{[value] \times 10^{-9}}{(1.5)^{3/2}} \left[\begin{array}{c}0\\4\\0\end{array}\right]\) and \(|\underline{E}|=\dfrac{1}{\pi\varepsilon_0}\ \dfrac{[value] \times 10^{-9}}{(1.5)^{3/2}}=\dfrac{[value] \times 10^{-9}}{\pi\times 8.854\times 10^{-12}(1.5)^{3/2}} =[value]\ \mathrm{V}\,\mathrm{m}^{-1}\). 3. Suppose the chosen edge to be used connects \\(A(0,0,1)\\) to \\(B(1,0,1)\\) then the centre point \\((G)\\) will be located at \\(G\left(\dfrac{1}{2}, 0, 1\right)\\). By symmetry, the field at \(G\) due to the charges at \(A\) and \(B\) will be zero. We note that the distances \(DG, \ OG, \ PG\) and \(SG\) are all equal. In the case of \(OG\) we calculate by Pythagoras that this distance is \(\sqrt{\left(\dfrac{1}{2}\right)^2+0^2+1^2}=\sqrt{1.25}\). Similarly the distances \(TG\) and \(RG\) are equal to \(\sqrt{2.25}\). Using the result that \(\underline{E}=\dfrac{1}{4\pi \varepsilon_0} \sum \dfrac{q_i\underline{r}_i}{|\underline{r}_i|^3}\) gives \(\qquad \underline{E}=\dfrac{[value] \times 10^{-9}}{4\pi\varepsilon_0}\left[\dfrac{1}{(1.25)^{3/2}}\left\{\left[\begin{array}{c}-\dfrac{1}{2}\\0\\1\end{array}\right]+\left[\begin{array}{c}\dfrac{1}{2}\\0\\1\end{array}\right]+\left[\begin{array}{c}-\dfrac{1}{2}\\-1\\0\end{array}\right]+\left[\begin{array}{c}\dfrac{1}{2}\\-1\\0\end{array}\right]\right\}\right.\) \(\qquad\qquad\qquad\left.+\dfrac{1}{(2.25)^{3/2}}\left\{\left[\begin{array}{c}-\dfrac{1}{2}\\-1\\1\end{array}\right]+\left[\begin{array}{c}\dfrac{1}{2}\\-1\\1\end{array}\right]\right\}\right]\) \(\qquad\quad =\dfrac{[value] \times 10^{-9}}{4\pi\varepsilon_0}\left[\dfrac{1}{(1.25)^{3/2}}\left[\begin{array}{c}0\\-2\\2\end{array}\right]+\dfrac{1}{(2.25)^{3/2}}\left[\begin{array}{c}0\\-2\\2\end{array}\right]\right]\) \(\qquad\quad =\dfrac{[value] \times 10^{-9}}{4\pi \varepsilon_0}\left[\begin{array}{c}0\\-2.02367\\2.02367\end{array}\right]\) Thus \(|\underline{E}|=\dfrac{[value] \times 10^{-9}}{4\times \pi\times 8.854\times 10^{-12}} \sqrt{0^2+(-2.02367)^2+(2.02367)^2}\) \(\qquad\qquad =[value]\ \mathrm{V}\,\mathrm{m}^{-1}\) (2 d.p.).