Question
What is the value and the direction of an electric field at a distance of 2.5 m
from a +1 nC
charge? Use 8.99×109 Nm2/C2
for Coulomb's constant.(1 point)
Responses
approximately 145 V/m
directed away from the positive charge
approximately 145 volt per meters directed away from the positive charge
approximately 145 V/m
directed toward the positive charge
approximately 145 volt per meters directed toward the positive charge
approximately 1.45 V/m
directed toward the positive charge
approximately 1 point 4 5 volt per meters directed toward the positive charge
approximately 1.45 V/m
directed away from the positive charge
from a +1 nC
charge? Use 8.99×109 Nm2/C2
for Coulomb's constant.(1 point)
Responses
approximately 145 V/m
directed away from the positive charge
approximately 145 volt per meters directed away from the positive charge
approximately 145 V/m
directed toward the positive charge
approximately 145 volt per meters directed toward the positive charge
approximately 1.45 V/m
directed toward the positive charge
approximately 1 point 4 5 volt per meters directed toward the positive charge
approximately 1.45 V/m
directed away from the positive charge
Answers
GPT-4o mini
To calculate the electric field (E) created by a point charge (Q) at a distance (r), we use the formula:
\[
E = \frac{k \cdot |Q|}{r^2}
\]
where:
- \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\),
- \(Q\) is the charge in coulombs,
- \(r\) is the distance from the charge in meters.
In this case:
- \(Q = +1 \, \text{nC} = +1 \times 10^{-9} \, \text{C}\),
- \(r = 2.5 \, \text{m}\).
Now, substituting the values into the formula:
\[
E = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 1 \times 10^{-9} \, \text{C}}{(2.5 \, \text{m})^2}
\]
Calculating the denominator:
\[
(2.5 \, \text{m})^2 = 6.25 \, \text{m}^2
\]
Now substituting back:
\[
E = \frac{8.99 \times 10^9 \cdot 1 \times 10^{-9}}{6.25}
\]
\[
E = \frac{8.99}{6.25} \, \text{V/m}
\]
\[
E \approx 1.4384 \, \text{V/m} \approx 1.44 \, \text{V/m}
\]
Thus, the electric field is approximately \(1.44 \, \text{V/m}\).
Since the charge is positive, the electric field lines point away from the charge.
### Final Answer
The value and direction of the electric field are approximately **1.45 V/m directed away from the positive charge**.
So the correct response would be:
"approximately 1.45 V/m directed away from the positive charge" or "approximately 1 point 4 5 volt per meters directed away from the positive charge".
\[
E = \frac{k \cdot |Q|}{r^2}
\]
where:
- \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\),
- \(Q\) is the charge in coulombs,
- \(r\) is the distance from the charge in meters.
In this case:
- \(Q = +1 \, \text{nC} = +1 \times 10^{-9} \, \text{C}\),
- \(r = 2.5 \, \text{m}\).
Now, substituting the values into the formula:
\[
E = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 1 \times 10^{-9} \, \text{C}}{(2.5 \, \text{m})^2}
\]
Calculating the denominator:
\[
(2.5 \, \text{m})^2 = 6.25 \, \text{m}^2
\]
Now substituting back:
\[
E = \frac{8.99 \times 10^9 \cdot 1 \times 10^{-9}}{6.25}
\]
\[
E = \frac{8.99}{6.25} \, \text{V/m}
\]
\[
E \approx 1.4384 \, \text{V/m} \approx 1.44 \, \text{V/m}
\]
Thus, the electric field is approximately \(1.44 \, \text{V/m}\).
Since the charge is positive, the electric field lines point away from the charge.
### Final Answer
The value and direction of the electric field are approximately **1.45 V/m directed away from the positive charge**.
So the correct response would be:
"approximately 1.45 V/m directed away from the positive charge" or "approximately 1 point 4 5 volt per meters directed away from the positive charge".