\( k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2 \)
a) We will find the electric field at P due to \(+3.0 \, \mu\text{C}\) at \(\rm B (0, 4.0 \, m) \). The values are given below.
Charge \( q = +3.0 \, \mu\text{C} = +3.0 \times 10^{-6} \, \text{C} \)
Position of \( q \): \( (0,\, 4.0 \, \text{m}) \)
Point \(\rm P \): \( (2.0 \, \text{m},\,0) \)
Calculate distance (\( r \)) between \( q \) and \(\rm P\)
\(r = \sqrt{(2.0 - 0)^2 + (0 - 4.0)^2} = 2\sqrt{5} \, \text{m}\)
Apply Coulomb’s Law for electric field magnitude
\(E = k \frac{|q|}{r^2} = 1.35 \times 10^3 \, \text{N/C}.\)
Determine direction (optional): Since \( q \) is \(positive\), the field points away from \( q \). We have a vector direction from \( \rm B(0, 4) \) to \( \rm P(2, 0) \) (downward and rightward). Angle \( \theta \) below \(+x\)-axis:
\(\tan \theta = \frac{4}{2} \implies \theta = \tan^{-1}(2) \approx 63.4^\circ .\)
Finally, we should write
\(\boxed{1.35 \times 10^3 \, \text{N/C}\, \text{at}\, 63.4^\circ\, \text{below the +x-axis}.}\)
b) We will find the electric field at \(\rm P\) due to \(−3.0\,\mu\rm C\) at \(\rm A(1\, m,\, 0)\),
Charge \( q = -3.0 \, \mu\text{C} = -3.0 \times 10^{-6} \, \text{C} \) ,
Position of \( q\): \( (1.0 \, \text{m},\,0) \),
Point \( \rm P \): \( (2.0 \, \text{m},\, 0) \) .
Calculate distance (\( r \)):
\(r = 2.0 \, \text{m} - 1.0 \, \text{m} = 1.0 \, \text{m}.\)
We apply Coulomb’s Law for magnitude
\(E = k \frac{|q|}{r^2} = 27 \times 10^3 \, \text{N/C}.\)
Determine direction (optional): Since \( q \) is \(negative\), the field points toward the charge. We find the field direction along the \(−x\) axis (from \(\rm P\) to \(\rm A\)). Finally, we must write
\(\boxed{2.7 \times 10^4 \, \text{N/C} \ \text{in the} \ -\hat{x} \ \text{direction}.}\)
We can easily apply Coulomb's Law:
\(F = k \frac{|q_1|.|q_2|}{r^2} = 60.0 \, \text{N} \).
The acceleration is found as
\(a = \frac{F}{m} = \frac{60}{1.20} = 50 \, \text{m/s}^2\)
Correct Answer: \(\boxed{B}\)