A +5.0-µC point charge is placed at the 0 cm mark of a meter stick and a -4.0-µC charge is placed at the 50 cm mark. What is the net electric field at the 30 cm mark? (k = 1/4pe0 = 8.99 × 10^9 N m²/C²)
91) __
A) 9.0 × 10^5 N/C
B) 4.0 × 10^5 N/C
C) 1.4 × 10^6 N/C
D) 5.0 × 10^5 N/C
The correct answer and explanation is :
To find the net electric field at the 30 cm mark due to the two point charges placed at the 0 cm and 50 cm marks, we need to apply Coulomb’s law for electric fields and consider the contributions of both charges.
Step 1: Formula for the Electric Field
The electric field (E) at a point due to a point charge is given by:
[
E = \frac{k |Q|}{r^2}
]
where:
- (E) is the electric field,
- (k = 8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2) is Coulomb’s constant,
- (|Q|) is the magnitude of the charge,
- (r) is the distance from the charge to the point where the field is being calculated.
Step 2: Electric Field Due to the +5.0-µC Charge at the 0 cm Mark
The distance between the 0 cm mark (where the +5.0-µC charge is located) and the 30 cm mark is 30 cm = 0.30 m. The electric field due to this charge is:
[
E_1 = \frac{k \times |Q_1|}{r_1^2} = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-6}}{(0.30)^2}
]
[
E_1 = \frac{8.99 \times 10^9 \times 5.0 \times 10^{-6}}{0.09} = 4.99 \times 10^5 \, \text{N/C}
]
The direction of the electric field due to a positive charge is away from the charge. Since the +5.0-µC charge is to the left of the 30 cm mark, the electric field at the 30 cm mark will point to the right.
Step 3: Electric Field Due to the -4.0-µC Charge at the 50 cm Mark
The distance between the 50 cm mark (where the -4.0-µC charge is located) and the 30 cm mark is 20 cm = 0.20 m. The electric field due to this charge is:
[
E_2 = \frac{k \times |Q_2|}{r_2^2} = \frac{8.99 \times 10^9 \times 4.0 \times 10^{-6}}{(0.20)^2}
]
[
E_2 = \frac{8.99 \times 10^9 \times 4.0 \times 10^{-6}}{0.04} = 8.99 \times 10^5 \, \text{N/C}
]
The direction of the electric field due to a negative charge is toward the charge. Since the -4.0-µC charge is to the right of the 30 cm mark, the electric field at the 30 cm mark will point to the left.
Step 4: Net Electric Field
The net electric field is the vector sum of the fields due to both charges. Since the fields due to the +5.0-µC charge and the -4.0-µC charge are in opposite directions, we subtract the magnitudes of the fields:
[
E_{\text{net}} = E_2 – E_1 = 8.99 \times 10^5 – 4.99 \times 10^5 = 4.0 \times 10^5 \, \text{N/C}
]
Final Answer:
The net electric field at the 30 cm mark is 4.0 × 10^5 N/C.
Thus, the correct answer is B) 4.0 × 10^5 N/C.