Calculate the magnitude of the electric field 2.47 m from a point charge of 4.79 mC.
The correct answer and explanation is:
To calculate the magnitude of the electric field created by a point charge, we use Coulomb’s Law for electric fields: E=k⋅∣q∣r2E = \frac{k \cdot |q|}{r^2}
Where:
- EE is the electric field (in N/C),
- kk is Coulomb’s constant: 8.99×109 Nm2/C28.99 \times 10^9 \, \text{Nm}^2/\text{C}^2,
- ∣q∣|q| is the magnitude of the point charge (in Coulombs),
- rr is the distance from the charge (in meters).
Given:
- q=4.79 mC=4.79×10−3 Cq = 4.79 \, \text{mC} = 4.79 \times 10^{-3} \, \text{C},
- r=2.47 mr = 2.47 \, \text{m}
Step-by-step Calculation:
E=8.99×109⋅4.79×10−3(2.47)2E = \frac{8.99 \times 10^9 \cdot 4.79 \times 10^{-3}}{(2.47)^2}
First, square the distance: (2.47)2=6.1009(2.47)^2 = 6.1009
Now calculate the numerator: 8.99×109⋅4.79×10−3=4.30521×1078.99 \times 10^9 \cdot 4.79 \times 10^{-3} = 4.30521 \times 10^7
Now divide: E=4.30521×1076.1009≈7.055×106 N/CE = \frac{4.30521 \times 10^7}{6.1009} \approx 7.055 \times 10^6 \, \text{N/C}
✅ Final Answer:
7.06×106 N/C\boxed{7.06 \times 10^6 \, \text{N/C}}
📘 Explanation (300 words):
The electric field (EE) is a measure of the force per unit charge that would be experienced by a small test charge placed at a certain point in space. For a point charge, this field radiates outward (if the charge is positive) or inward (if negative) and decreases with the square of the distance from the charge, following an inverse square law.
In this case, we have a point charge of 4.79 mC (milliCoulombs), which we convert to Coulombs by multiplying by 10−310^{-3}, giving us 4.79×10−3 C4.79 \times 10^{-3} \, \text{C}. The point where we want to measure the electric field is 2.47 meters away from the charge.
Using the formula E=k⋅∣q∣r2E = \frac{k \cdot |q|}{r^2}, we plug in the values: Coulomb’s constant k=8.99×109 Nm2/C2k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2, the charge q=4.79×10−3 Cq = 4.79 \times 10^{-3} \, \text{C}, and the distance r=2.47 mr = 2.47 \, \text{m}. After calculating the numerator and denominator separately, we divide them to get the magnitude of the electric field.
The result is approximately 7.06×106 N/C7.06 \times 10^6 \, \text{N/C}, which means any 1-Coulomb test charge placed 2.47 meters from this point charge would experience a force of about 7.06 million newtons. This demonstrates how strong electric forces can be, even at a few meters away, especially when the source charge is large.