Problem 1 calculates electric field strength using force and charge; Problem 2 involves finding the electric field at the origin due to two point charges in a Cartesian plane.

Problem 1: Magnitude of the Electric Field Strength

#### Given (G):
- Charge, \( q = 15 \, \mu\text{C} = 15 \times 10^{-6} \, \text{C} \)
- Force experienced by the charge, \( F = 599 \, \text{N} \)

#### Required (R):
- Magnitude of the electric field strength, \( E \)

#### Equation (E):
The relationship between force, charge, and electric field is given by:
\[
F = qE
\]
Rearranging for \( E \):
\[
E = \frac{F}{q}
\]

#### Solution (S):
Substitute the given values into the equation:
\[
E = \frac{599 \, \text{N}}{15 \times 10^{-6} \, \text{C}}
\]
Perform the division:
\[
E = \frac{599}{15 \times 10^{-6}} = \frac{599}{1.5 \times 10^{-5}} = 39933333.33 \, \text{N/C}
\]
Express the result in scientific notation with appropriate significant figures:
\[
E \approx 4.0 \times 10^7 \, \text{N/C}
\]

#### Answer (A):
The magnitude of the electric field strength is:
\[
\boxed{4.0 \times 10^7 \, \text{N/C}}
\]

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Problem 2: Electric Field at the Origin Due to Two Charges

#### Given (G):
- Charge \( q_1 = -5.4 \, \text{nC} = -5.4 \times 10^{-9} \, \text{C} \) located at \( (8.0 \, \text{m}, 0) \)
- Charge \( q_2 = 8.8 \, \text{nC} = 8.8 \times 10^{-9} \, \text{C} \) located at \( (0, 5.0 \, \text{m}) \)

#### Required (R):
- Find the electric field at the origin due to both charges.
- Draw a free-body diagram and identify the directions of \( \vec{E}_1 \) and \( \vec{E}_2 \).

#### Equation (E):
The electric field due to a point charge \( q \) at a distance \( r \) is given by:
\[
\vec{E} = k \frac{q}{r^2} \hat{r}
\]
where \( k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \) is Coulomb's constant, and \( \hat{r} \) is the unit vector in the direction from the charge to the point where the field is being calculated.

#### Solution (S):

1. Electric Field Due to \( q_1 \):
- Position of \( q_1 \): \( (8.0 \, \text{m}, 0) \)
- Distance from \( q_1 \) to the origin:
\[
r_1 = \sqrt{(8.0)^2 + 0^2} = 8.0 \, \text{m}
\]
- Unit vector \( \hat{r}_1 \) from \( q_1 \) to the origin:
\[
\hat{r}_1 = -\hat{i}
\]
- Electric field \( \vec{E}_1 \):
\[
\vec{E}_1 = k \frac{q_1}{r_1^2} \hat{r}_1 = 8.99 \times 10^9 \frac{-5.4 \times 10^{-9}}{(8.0)^2} (-\hat{i})
\]
Simplify:
\[
\vec{E}_1 = 8.99 \times 10^9 \cdot \frac{-5.4 \times 10^{-9}}{64} \cdot (-\hat{i}) = 8.99 \cdot \frac{5.4}{64} \cdot 10^0 \hat{i}
\]
\[
\vec{E}_1 = 8.99 \cdot 0.084375 \hat{i} = 0.758 \hat{i} \, \text{N/C}
\]

2. Electric Field Due to \( q_2 \):
- Position of \( q_2 \): \( (0, 5.0 \, \text{m}) \)
- Distance from \( q_2 \) to the origin:
\[
r_2 = \sqrt{0^2 + (5.0)^2} = 5.0 \, \text{m}
\]
- Unit vector \( \hat{r}_2 \) from \( q_2 \) to the origin:
\[
\hat{r}_2 = -\hat{j}
\]
- Electric field \( \vec{E}_2 \):
\[
\vec{E}_2 = k \frac{q_2}{r_2^2} \hat{r}_2 = 8.99 \times 10^9 \frac{8.8 \times 10^{-9}}{(5.0)^2} (-\hat{j})
\]
Simplify:
\[
\vec{E}_2 = 8.99 \times 10^9 \cdot \frac{8.8 \times 10^{-9}}{25} \cdot (-\hat{j}) = 8.99 \cdot \frac{8.8}{25} \cdot 10^0 (-\hat{j})
\]
\[
\vec{E}_2 = 8.99 \cdot 0.352 \cdot (-\hat{j}) = -3.16 \hat{j} \, \text{N/C}
\]

3. Total Electric Field at the Origin:
- Add the contributions from \( \vec{E}_1 \) and \( \vec{E}_2 \):
\[
\vec{E}_{\text{total}} = \vec{E}_1 + \vec{E}_2 = 0.758 \hat{i} - 3.16 \hat{j} \, \text{N/C}
\]

#### Free-Body Diagram:
- Draw the Cartesian plane with the origin at the center.
- Place \( q_1 \) at \( (8.0, 0) \) and \( q_2 \) at \( (0, 5.0) \).
- Draw arrows representing \( \vec{E}_1 \) pointing towards the left (since \( q_1 \) is negative) and \( \vec{E}_2 \) pointing downwards (since \( q_2 \) is positive).

#### Answer (A):
The total electric field at the origin is:
\[
\boxed{0.758 \hat{i} - 3.16 \hat{j} \, \text{N/C}}
\]