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 x 10^9 Nm^2/C^2 for Coulomb's constant.
answers:
approximately 145 V/m directed away from the positive charge

approximately 1.45 V/m directed toward the positive charge

approximately 1.45 V/m directed away from the positive charge

approximately 145 V/m directed toward the positive charge

Answer 1

To calculate the electric field (E) at a distance (r) from a point charge (Q), you can use the formula for the electric field created by a point charge:

\[
E = \frac{k \cdot |Q|}{r^2}
\]

where:
- \(E\) is the electric field in volts per meter (V/m),
- \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\),
- \(|Q|\) is the magnitude of the charge in coulombs (C),
- \(r\) is the distance from the charge in meters (m).

Given:
- \(Q = +1 \, \text{nC} = 1 \times 10^{-9} \, \text{C}\)
- \(r = 2.5 \, \text{m}\)

Now, substitute the values into the electric field formula:

\[
E = \frac{8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \cdot 1 \times 10^{-9} \, \text{C}}{(2.5 \, \text{m})^2}
\]

Calculating the denominator:
\[
(2.5 \, \text{m})^2 = 6.25 \, \text{m}^2
\]

Now, calculate \(E\):
\[
E = \frac{8.99 \times 10^9 \cdot 1 \times 10^{-9}}{6.25}
\]
\[
E = \frac{8.99}{6.25} \, \text{V/m}
\]
\[
E \approx 1.4384 \, \text{V/m}
\]

Now we can approximate this to:
\[
E \approx 1.44 \, \text{V/m}
\]

The direction of the electric field produced by a positive charge is away from the charge.

Therefore, the final answer is:
**approximately 1.45 V/m directed away from the positive charge**.