Answer :
We wish to find the electric field intensity at the point
[tex]$$
(3,4) \text{ m}
$$[/tex]
due to three point charges:
- [tex]$Q_1 = +5 \times 10^{-8}$[/tex] C located at [tex]$(0,0)$[/tex],
- [tex]$Q_2 = +4 \times 10^{-8}$[/tex] C located at [tex]$(3,0)$[/tex], and
- [tex]$Q_3 = -6 \times 10^{-8}$[/tex] C located at [tex]$(0,4)$[/tex].
The electric field [tex]$\mathbf{E}$[/tex] from a point charge is given by
[tex]$$
\mathbf{E} = k \frac{Q}{r^2} \hat{r},
$$[/tex]
where
- [tex]$k = 8.987551787\times10^9\ \text{N m}^2/\text{C}^2$[/tex] is Coulomb's constant,
- [tex]$r$[/tex] is the distance from the charge to the field point,
- [tex]$\hat{r}$[/tex] is a unit vector directed from the charge to the field point.
We will calculate the contribution from each charge separately.
──────────────────────────────
Step 1. Electric Field Due to [tex]$Q_1$[/tex]
1.1. Determine the displacement from [tex]$Q_1$[/tex] at [tex]$(0,0)$[/tex] to the field point [tex]$(3,4)$[/tex]:
[tex]$$
\text{disp}_1 = (3-0,\, 4-0) = (3,4).
$$[/tex]
1.2. Compute the distance:
[tex]$$
r_1 = \sqrt{3^2+4^2} = 5\ \text{m}.
$$[/tex]
1.3. Determine the unit vector:
[tex]$$
\hat{r}_1 = \left(\frac{3}{5}, \frac{4}{5}\right) = (0.6,\, 0.8).
$$[/tex]
1.4. Calculate the magnitude of the electric field:
[tex]$$
E_1 = k \frac{Q_1}{r_1^2} = 8.987551787\times10^9 \cdot \frac{5 \times 10^{-8}}{25}.
$$[/tex]
Multiplying out and then including the unit vector gives
[tex]$$
\mathbf{E}_1 \approx (10.7851,\ 14.3801)\ \text{N/C}.
$$[/tex]
──────────────────────────────
Step 2. Electric Field Due to [tex]$Q_2$[/tex]
2.1. Displacement from [tex]$Q_2$[/tex] at [tex]$(3,0)$[/tex] to the field point [tex]$(3,4)$[/tex]:
[tex]$$
\text{disp}_2 = (3-3,\, 4-0) = (0,4).
$$[/tex]
2.2. Distance:
[tex]$$
r_2 = \sqrt{0^2+4^2} = 4\ \text{m}.
$$[/tex]
2.3. Unit vector:
[tex]$$
\hat{r}_2 = \left(0,\, 1\right).
$$[/tex]
2.4. Electric field magnitude:
[tex]$$
E_2 = k \frac{Q_2}{r_2^2} = 8.987551787\times10^9 \cdot \frac{4 \times 10^{-8}}{16}.
$$[/tex]
Thus, the vector is
[tex]$$
\mathbf{E}_2 \approx (0,\ 22.4689)\ \text{N/C}.
$$[/tex]
──────────────────────────────
Step 3. Electric Field Due to [tex]$Q_3$[/tex]
3.1. Displacement from [tex]$Q_3$[/tex] at [tex]$(0,4)$[/tex] to the field point [tex]$(3,4)$[/tex]:
[tex]$$
\text{disp}_3 = (3-0,\, 4-4) = (3,0).
$$[/tex]
3.2. Distance:
[tex]$$
r_3 = \sqrt{3^2+0^2} = 3\ \text{m}.
$$[/tex]
3.3. Unit vector:
[tex]$$
\hat{r}_3 = (1,\, 0).
$$[/tex]
3.4. Since [tex]$Q_3$[/tex] is negative, its electric field will point toward the charge. However, when computing with the given formula the sign of [tex]$Q_3$[/tex] takes care of the direction. Its magnitude is
[tex]$$
E_3 = k \frac{|Q_3|}{r_3^2} = 8.987551787\times10^9 \cdot \frac{6 \times 10^{-8}}{9}.
$$[/tex]
Including the negative sign and the unit vector direction gives
[tex]$$
\mathbf{E}_3 \approx (-59.9170,\, 0)\ \text{N/C}.
$$[/tex]
──────────────────────────────
Step 4. Find the Total Electric Field
The electric field is a vector quantity, so we sum the components from each charge:
4.1. Sum of the [tex]$x$[/tex]-components:
[tex]$$
E_{x,\text{total}} = 10.7851 + 0 - 59.9170 \approx -49.1319\ \text{N/C}.
$$[/tex]
4.2. Sum of the [tex]$y$[/tex]-components:
[tex]$$
E_{y,\text{total}} = 14.3801 + 22.4689 + 0 \approx 36.8490\ \text{N/C}.
$$[/tex]
Thus, the total electric field vector is:
[tex]$$
\mathbf{E}_{\text{total}} = (-49.1319,\ 36.8490)\ \text{N/C}.
$$[/tex]
──────────────────────────────
Step 5. Determine the Magnitude of the Total Electric Field
The magnitude is calculated by:
[tex]$$
|\mathbf{E}_{\text{total}}| = \sqrt{(-49.1319)^2 + (36.8490)^2} \approx 61.4149\ \text{N/C}.
$$[/tex]
──────────────────────────────
Final Answer
The electric field intensity at the point [tex]$(3,4)$[/tex] is:
[tex]$$
\mathbf{E}_{\text{total}} \approx (-49.13,\ 36.85)\ \text{N/C}
$$[/tex]
with a magnitude of approximately
[tex]$$
61.41\ \text{N/C}.
$$[/tex]
[tex]$$
(3,4) \text{ m}
$$[/tex]
due to three point charges:
- [tex]$Q_1 = +5 \times 10^{-8}$[/tex] C located at [tex]$(0,0)$[/tex],
- [tex]$Q_2 = +4 \times 10^{-8}$[/tex] C located at [tex]$(3,0)$[/tex], and
- [tex]$Q_3 = -6 \times 10^{-8}$[/tex] C located at [tex]$(0,4)$[/tex].
The electric field [tex]$\mathbf{E}$[/tex] from a point charge is given by
[tex]$$
\mathbf{E} = k \frac{Q}{r^2} \hat{r},
$$[/tex]
where
- [tex]$k = 8.987551787\times10^9\ \text{N m}^2/\text{C}^2$[/tex] is Coulomb's constant,
- [tex]$r$[/tex] is the distance from the charge to the field point,
- [tex]$\hat{r}$[/tex] is a unit vector directed from the charge to the field point.
We will calculate the contribution from each charge separately.
──────────────────────────────
Step 1. Electric Field Due to [tex]$Q_1$[/tex]
1.1. Determine the displacement from [tex]$Q_1$[/tex] at [tex]$(0,0)$[/tex] to the field point [tex]$(3,4)$[/tex]:
[tex]$$
\text{disp}_1 = (3-0,\, 4-0) = (3,4).
$$[/tex]
1.2. Compute the distance:
[tex]$$
r_1 = \sqrt{3^2+4^2} = 5\ \text{m}.
$$[/tex]
1.3. Determine the unit vector:
[tex]$$
\hat{r}_1 = \left(\frac{3}{5}, \frac{4}{5}\right) = (0.6,\, 0.8).
$$[/tex]
1.4. Calculate the magnitude of the electric field:
[tex]$$
E_1 = k \frac{Q_1}{r_1^2} = 8.987551787\times10^9 \cdot \frac{5 \times 10^{-8}}{25}.
$$[/tex]
Multiplying out and then including the unit vector gives
[tex]$$
\mathbf{E}_1 \approx (10.7851,\ 14.3801)\ \text{N/C}.
$$[/tex]
──────────────────────────────
Step 2. Electric Field Due to [tex]$Q_2$[/tex]
2.1. Displacement from [tex]$Q_2$[/tex] at [tex]$(3,0)$[/tex] to the field point [tex]$(3,4)$[/tex]:
[tex]$$
\text{disp}_2 = (3-3,\, 4-0) = (0,4).
$$[/tex]
2.2. Distance:
[tex]$$
r_2 = \sqrt{0^2+4^2} = 4\ \text{m}.
$$[/tex]
2.3. Unit vector:
[tex]$$
\hat{r}_2 = \left(0,\, 1\right).
$$[/tex]
2.4. Electric field magnitude:
[tex]$$
E_2 = k \frac{Q_2}{r_2^2} = 8.987551787\times10^9 \cdot \frac{4 \times 10^{-8}}{16}.
$$[/tex]
Thus, the vector is
[tex]$$
\mathbf{E}_2 \approx (0,\ 22.4689)\ \text{N/C}.
$$[/tex]
──────────────────────────────
Step 3. Electric Field Due to [tex]$Q_3$[/tex]
3.1. Displacement from [tex]$Q_3$[/tex] at [tex]$(0,4)$[/tex] to the field point [tex]$(3,4)$[/tex]:
[tex]$$
\text{disp}_3 = (3-0,\, 4-4) = (3,0).
$$[/tex]
3.2. Distance:
[tex]$$
r_3 = \sqrt{3^2+0^2} = 3\ \text{m}.
$$[/tex]
3.3. Unit vector:
[tex]$$
\hat{r}_3 = (1,\, 0).
$$[/tex]
3.4. Since [tex]$Q_3$[/tex] is negative, its electric field will point toward the charge. However, when computing with the given formula the sign of [tex]$Q_3$[/tex] takes care of the direction. Its magnitude is
[tex]$$
E_3 = k \frac{|Q_3|}{r_3^2} = 8.987551787\times10^9 \cdot \frac{6 \times 10^{-8}}{9}.
$$[/tex]
Including the negative sign and the unit vector direction gives
[tex]$$
\mathbf{E}_3 \approx (-59.9170,\, 0)\ \text{N/C}.
$$[/tex]
──────────────────────────────
Step 4. Find the Total Electric Field
The electric field is a vector quantity, so we sum the components from each charge:
4.1. Sum of the [tex]$x$[/tex]-components:
[tex]$$
E_{x,\text{total}} = 10.7851 + 0 - 59.9170 \approx -49.1319\ \text{N/C}.
$$[/tex]
4.2. Sum of the [tex]$y$[/tex]-components:
[tex]$$
E_{y,\text{total}} = 14.3801 + 22.4689 + 0 \approx 36.8490\ \text{N/C}.
$$[/tex]
Thus, the total electric field vector is:
[tex]$$
\mathbf{E}_{\text{total}} = (-49.1319,\ 36.8490)\ \text{N/C}.
$$[/tex]
──────────────────────────────
Step 5. Determine the Magnitude of the Total Electric Field
The magnitude is calculated by:
[tex]$$
|\mathbf{E}_{\text{total}}| = \sqrt{(-49.1319)^2 + (36.8490)^2} \approx 61.4149\ \text{N/C}.
$$[/tex]
──────────────────────────────
Final Answer
The electric field intensity at the point [tex]$(3,4)$[/tex] is:
[tex]$$
\mathbf{E}_{\text{total}} \approx (-49.13,\ 36.85)\ \text{N/C}
$$[/tex]
with a magnitude of approximately
[tex]$$
61.41\ \text{N/C}.
$$[/tex]
