# Question 82085

Nov 2, 2016

Let charges be ${q}_{1} = + 10 \mu C$, ${q}_{2} = + 8 \mu C$ and ${q}_{3} = + 15 \mu C$ be placed as shown in the figure above. All three vertex angles $= {60}^{\circ}$ and all sides $a = 10 c m$
Now according to Coulomb's Law
${\vec{F}}_{13} = {k}_{e} \frac{{q}_{1} {q}_{3}}{r} _ {13}^{2} {\hat{r}}_{13}$

Inserting given values we get
$| {\vec{F}}_{13} | = \left(9.0 \times {10}^{9}\right) \frac{\left(10 \times {10}^{-} 6\right) \left(15 \times {10}^{-} 6\right)}{0.1} ^ 2 = 135 N$

Similarly
${\vec{F}}_{23} = {k}_{e} \frac{{q}_{2} {q}_{3}}{r} _ {23}^{2} {\hat{r}}_{23}$
From the figure we see that ${\hat{r}}_{23} = \hat{i}$
Inserting given values we get
$| {\vec{F}}_{23} | = \left(9.0 \times {10}^{9}\right) \frac{\left(8 \times {10}^{-} 6\right) \left(15 \times {10}^{-} 6\right)}{0.1} ^ 2 = 108 N$

From the figure it is clear that angle between the two unit vectors ${\hat{r}}_{13} \mathmr{and} \hat{i}$ is ${60}^{\circ}$
As such the $x \mathmr{and} y$ components of ${\vec{F}}_{13}$ can be written as
${\vec{F}}_{13} = 135 \cos {60}^{\circ} \hat{i} - 135 \sin {60}^{\circ} \hat{j}$
${\vec{F}}_{13} = 67.5 \hat{i} - 116.9 \hat{j}$
Total force is sum of two forces
${\vec{F}}_{13} + {\vec{F}}_{23} = 67.5 \hat{i} - 116.9 \hat{j} + 108 \hat{i}$
${\vec{F}}_{\text{total}} = \left(175.5 \hat{i} - 116.9 \hat{j}\right) N$

Nov 9, 2016

$\textsf{211 \textcolor{w h i t e}{x} N}$ at an angle of $\textsf{{56}^{\circ}}$ to the vertical.

#### Explanation:

I will suggest two methods you could use which involves finding the resultant of the two forces acting on the charge.

Here is the situation:

$\textsf{{q}_{1} = + 10 \mu C}$

$\textsf{{q}_{2} = + 8 \mu C}$

$\textsf{{q}_{3} = + 15 \mu C}$

Method (1)

Here is the vector diagram showing the forces acting on $\textsf{{q}_{3}}$:

There is a horizontal force sf(stackrel(rarr)(F_(23)). There is a force sf(stackrel(rarr)(F_(13)) at an angle of $\textsf{{60}^{\circ}}$ to the horizontal.

These are indicated by the red arrows. We need to get the resultant R.

Coulombs Law gives us the force of attraction between $\textsf{{q}_{2}}$ and $\textsf{{q}_{3}}$:

$\textsf{\stackrel{\rightarrow}{{F}_{23}} = \frac{k {q}_{2} {q}_{3}}{{r}_{23}^{2}} . \hat{r}}$

$\textsf{k}$ is a constant with the value $\textsf{9 \times {10}^{9} \text{ } N . {m}^{2} . {C}^{- 2}}$

$\therefore$$\textsf{\stackrel{\rightarrow}{{F}_{23}} = \frac{k \times 8 \times {10}^{- 6} \times 15 \times {10}^{- 6}}{{0.1}^{2}} = 120 k \times {10}^{- 10} \text{ } N}$

By the same reasoning:

$\textsf{\stackrel{\rightarrow}{{F}_{13}} = \frac{k \times 10 \times {10}^{- 6} \times 15 \times {10}^{- 6}}{{0.1}^{2}} = 150 k \times {10}^{- 10} \text{ } N}$

To make the numbers easier to handle I will normalise the forces in units of $\textsf{K}$ where $\textsf{K = k \times {10}^{- 10} \text{ } N}$

Now we have a Side Angle Side triangle. We know 2 sides and the angle between them so we can apply the cosine rule to find the unknown side.

For a triangle abc this gives us:

$\textsf{{a}^{2} = {b}^{2} + {c}^{2} - \text{2bc} C o s A}$

Applying this to the vector diagram we get:

sf(R^2=150^2+120^2-(2xx150xx120xxcos120)

$\textsf{{R}^{2} = 22500 + 14400 - \left(- 18000\right)}$

$\textsf{{R}^{2} = 54900}$

$\textsf{R = \sqrt{54900} = 234.3 \text{ } K}$ units

Re-scaling:

$\textsf{R = 234.3 \times 9 \times {10}^{9} \times {10}^{- 10} = \textcolor{red}{211} \text{ } N}$

To find the angle $\textsf{\theta}$ we can apply the sine rule:

$\textsf{\frac{234}{\sin} 120 = \frac{120}{\sin} \theta}$

$\therefore$$\textsf{\sin \theta = 0.444}$

From which:

$\textsf{\theta = {26.3}^{\circ}}$

From the diagram you can see that resultant R must be at an angle of $\textsf{{30}^{\circ} + {26.3}^{\circ} = {56.3}^{\circ}}$ to the vertical.

Method (2)

We can resolve the forces into their horizontal and vertical components and find the resultant from that.

Horizontal components:

The horizontal component of $\textsf{\stackrel{\rightarrow}{{F}_{13}} = \textsf{\stackrel{\rightarrow}{{F}_{13}} \cos 60} = \stackrel{\rightarrow}{{F}_{13}} \times 0.5}$

So the total horizontal component is given by:

$\textsf{{F}_{x} = \stackrel{\rightarrow}{{F}_{23}} + 0.5 \times \stackrel{\rightarrow}{{F}_{13}}}$

$\textsf{{F}_{x} = 120 K + 0.5 x 150 K = 195 K}$

Vertical components:

sf(stackrel(rarr)(F_(23))# does not have a vertical component so:

$\textsf{{F}_{y} = \stackrel{\rightarrow}{{F}_{13}} \cos 30 = 150 K \times 0.866 = 129.9 K}$

Now we can apply Pythagoras:

$\textsf{{\left(129.9 K\right)}^{2} + {\left(195 K\right)}^{2} = {R}^{2}}$

$\textsf{{R}^{2} = 54899 {K}^{2}}$

$\textsf{R = 234.3 K \text{ } N}$

$\textsf{R = 234.3 \times 9 \times {10}^{9} \times {10}^{- 10} = \textcolor{red}{211} \text{ } N}$

To find the angle $\alpha$ to the vertical:

$\textsf{T a n \alpha = \frac{195}{129.9} = 1.5}$

From which:

$\alpha = {56.3}^{\circ}$

So the 2 methods are in agreement.