1. Concept: Electric field in vector form is the product of its magnitude and the unit vector in the direction of the position vector.
2. Formula: $$ \vec{E} = \frac{k q}{r^2} \hat{r} = \frac{k q}{r^3} \vec{r} $$
3. Solution: - Calculate magnitude of position vector: \( r = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ m} \).
- Calculate the scalar magnitude of the electric field: \( E = \frac{9 \times 10^9 \times 5 \times 10^{-6}}{5^2} = \frac{45000}{25} = 1800 \text{ V/m} \).
- Find the unit vector: \( \hat{r} = \frac{4\hat{i} + 3\hat{j}}{5} \).
- Multiply magnitude by unit vector: \( \vec{E} = 1800 \left( \frac{4\hat{i} + 3\hat{j}}{5} \right) \).
- Simplify: \( \vec{E} = 360 (4\hat{i} + 3\hat{j}) = 1440\hat{i} + 1080\hat{j} \text{ V/m} \).
4. Why other options are incorrect: Option C uses incorrect units for electric field (N/m instead of V/m or N/C). Options A and D arise from arithmetic failures during vector component multiplication.
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