20.2 Calculate the net electrostatio force that $$ A $$ and $$ B $$ exert on $$ C $$.

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We will illustrate a complete solution by assuming the following configuration. (In your particular textbook the numbers and positions may differ. In our example the values below are assumed.)

Assume that three point charges are located at the vertices of an isosceles “horizontal‐base” triangle as follows:

• Charge $$ q_A = +3.0\,\mu\text{C} $$ is at point $$ A=(-1.0\,\text{m},0) $$.

• Charge $$ q_B = +3.0\,\mu\text{C} $$ is at point $$ B=(1.0\,\text{m},0) $$.

• Charge $$ q_C = -3.0\,\mu\text{C} $$ is at point $$ C=(0,1.0\,\text{m}) $$.

When a negative charge is attracted toward a positive charge the force on the negative charge points toward the positive charge. Thus the force on $$C$$ due to $$A$$ (denoted $$\vec{F}_{AC}$$) points from $$C$$ toward $$A$$, and the force on $$C$$ due to $$B$$ (denoted $$\vec{F}_{BC}$$) points from $$C$$ toward $$B$$.

Below are the step‐by‐step calculations.

---

### 1. Determine the Distances

The distance from $$C$$ to either $$A$$ or $$B$$ is given by the distance formula. For example, for $$ A $$ to $$ C $$:

$$
r_{AC} = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{(0 - (-1.0))^2 + (1.0-0)^2} = \sqrt{1^2+1^2} = \sqrt{2}\,.
$$

Similarly,

$$
r_{BC} = \sqrt{(0 - 1.0)^2 + (1.0-0)^2} = \sqrt{2}\,.
$$

---

### 2. Compute the Magnitudes of the Forces

Coulomb’s law gives the magnitude of the force between two point charges:

$$
F = k\,\frac{|q_iq_j|}{r^2}\,,
$$

where $$ k = 8.99\times10^9\,\text{N·m}^2/\text{C}^2 $$. First, express all charges in Coulombs:

$$
3.0\,\mu\text{C} = 3.0\times10^{-6}\,\text{C}\,.
$$

Since the distance squared is

$$
r^2 = (\sqrt{2})^2 = 2\,,
$$

the magnitude of the force exerted by $$A$$ on $$C$$ is

$$
F_{AC} = k\,\frac{|q_Aq_C|}{r_{AC}^2} = 8.99\times10^9\,\frac{(3.0\times10^{-6})(3.0\times10^{-6})}{2}\,.
$$

Calculate the product in the numerator:

$$
(3.0\times10^{-6})^2 = 9.0\times10^{-12}\,.
$$

Thus,

$$
F_{AC} = 8.99\times10^9\,\frac{9.0\times10^{-12}}{2} = 8.99\times10^9\,\times4.5\times10^{-12}\,.
$$

Multiply the numbers:

$$
8.99\times4.5 \approx 40.455\,,
$$

so

$$
F_{AC} \approx 40.455\times10^{-3}\,\text{N} = 0.0405\,\text{N}\,.
$$

Because of symmetry, the force from $$B$$ on $$C$$ has the same magnitude:

$$
F_{BC} = 0.0405\,\text{N}\,.
$$

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### 3. Determine the Directions and Components of the Forces

**Force $$\vec{F}_{AC}$$:**  
The force on $$C$$ due to $$A$$ points from $$C$$ at $$(0,1)$$ toward $$A$$ at $$(-1,0)$$. The vector from $$C$$ to $$A$$ is

$$
\vec{r}_{CA} = (x_A - x_C,\, y_A - y_C) = (-1-0,\, 0-1) = (-1,\,-1)\,.
$$

Its magnitude is

$$
|\vec{r}_{CA}| = \sqrt{(-1)^2+(-1)^2} = \sqrt{2}\,.
$$

Thus the unit vector in the direction from $$C$$ to $$A$$ is

$$
\hat{r}_{CA} = \left(\frac{-1}{\sqrt{2}},\,\frac{-1}{\sqrt{2}}\right)\,.
$$

So, the vector force is

$$
\vec{F}_{AC} = F_{AC}\,\hat{r}_{CA} = 0.0405\left(\frac{-1}{\sqrt{2}},\,\frac{-1}{\sqrt{2}}\right) \approx \left(-0.0287,\,-0.0287\right)\,\text{N}\,.
$$

**Force $$\vec{F}_{BC}$$:**  
The force on $$C$$ due to $$B$$ points from $$C$$ at $$(0,1)$$ toward $$B$$ at $$(1,0)$$. The vector from $$C$$ to $$B$$ is

$$
\vec{r}_{CB} = (1-0,\, 0-1) = (1,\,-1)\,.
$$

Its magnitude is also

$$
|\vec{r}_{CB}| = \sqrt{1^2+(-1)^2} = \sqrt{2}\,.
$$

Thus the unit vector is

$$
\hat{r}_{CB} = \left(\frac{1}{\sqrt{2}},\,\frac{-1}{\sqrt{2}}\right)\,.
$$

Then,

$$
\vec{F}_{BC} = F_{BC}\,\hat{r}_{CB} = 0.0405\left(\frac{1}{\sqrt{2}},\,\frac{-1}{\sqrt{2}}\right) \approx \left(0.0287,\,-0.0287\right)\,\text{N}\,.
$$

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### 4. Find the Net Force on $$C$$

The net force on $$C$$ is the vector sum

$$
\vec{F}_{\text{net}} = \vec{F}_{AC} + \vec{F}_{BC}\,.
$$

Adding the components:

- $$x$$-component:
  $$
  F_{\text{net},x} = -0.0287 + 0.0287 = 0\,.
  $$
- $$y$$-component:
  $$
  F_{\text{net},y} = -0.0287 + (-0.0287) = -0.0574\,\text{N}\,.
  $$

Thus,

\[
\vec{F}_{\text{net}} = \left(0
The net electrostatic force on $$ C $$ is $$ 0.0574\,\text{N} $$ downward.

20.2 Calculate the net electrostatio force that \( A \) and \( B \) exert on \( C \).

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