What is the value and the direction of an electric field at a distance of 2.5 m from a +1 nC charge?
approximately 145 V/m directed toward the positive charge
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
What forces do electric fields represent?
- They represent non-contact electrical forces.
What does it mean if electric field lines are close together versus farther apart?
- The strength of the electric field is greater.
What is the force on a 0.0050 C charge in an electric field of 300 V/m?
- 1.5 N
What is the strength of an electric field in which a charged particle of 3.0 × 10−9 C is 0.05 m from the test charge?
- 1.1 × 10⁴ N/C
What is the value and the direction of an electric field at a distance of 2.5 m from a +1 nC charge?
- approximately 1.45 V/m directed away from the positive charge
Hi Katy. The answer is D, approximately 1.45 V/m directed AWAY from the positive charge. Hope this helps.
nobody is still right as of february 23rd :) i hate physics
ty nobody
Well, isn't that a shocking question! Let me enlighten you with the answer. The value and direction of an electric field at a distance of 2.5 m from a +1 nC charge is approximately 1.45 V/m directed toward the positive charge. So, it's like the positive charge is saying, "Come to me, my electromagnetic friend!"
To determine the value and direction of an electric field at a given distance from a charge, you can use Coulomb's Law and the formula for calculating the electric field.
First, let's determine the value of the electric field. Coulomb's Law states that the electric field strength (E) created by a point charge (Q) at a distance (r) is given by the equation:
E = k * (Q / r^2)
where k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2).
In this case, the point charge (Q) is +1 nC (nanoCoulomb), and the distance (r) is 2.5 m.
Plugging these values into the equation, we get:
E = (9 x 10^9 Nm^2/C^2) * (1 x 10^-9 C) / (2.5 m)^2
Simplifying the equation, we have:
E = (9 x 1) / (2.5)^2 = 9 / 6.25 = 1.44 V/m
Therefore, the value of the electric field at a distance of 2.5 m from the +1 nC charge is approximately 1.44 V/m.
Now, let's determine the direction of the electric field. The electric field created by a positive charge points away from the charge, while the electric field created by a negative charge points towards the charge.
Since we have a positive charge, the electric field will be directed away from it. Therefore, the correct answer is:
approximately 1.45 V/m directed away from the positive charge