A charge of 4.0 nC is at +20 cm on the x-axis.Another charge, -3.0 nC, is at (+10 cm, -5 cm). Please show all work neatly and explain.
What is the electric field due to these two charges at the origin?
What is the force each charge exerts on the other?
Well, I must say, those charges seem to be getting quite cozy on the x-axis! Let's calculate the electric field and the force between them, shall we?
To find the electric field at the origin (0,0), we need to calculate the electric field produced by each charge separately and then add them up.
Let's start with the first charge, which is +4.0 nC located at +20 cm on the x-axis. The electric field produced by this charge can be found using the formula:
E1 = k * q1 / r1^2
where k is the electrostatic constant (k = 9 x 10^9 Nm^2/C^2), q1 is the charge (+4.0 nC = 4.0 x 10^-9 C), and r1 is the distance from the charge to the origin (r1 = 20 cm = 0.20 m).
Plugging in the values, we get:
E1 = (9 x 10^9 Nm^2/C^2) * (4.0 x 10^-9 C) / (0.20 m)^2
Calculating that should give you the electric field produced by the first charge at the origin.
Now, let's move on to the second charge, which is -3.0 nC located at (10 cm, -5 cm) on the coordinate plane. To find the electric field produced by this charge, we use the same formula:
E2 = k * q2 / r2^2
where q2 is the charge (-3.0 nC = -3.0 x 10^-9 C) and r2 is the distance from the charge to the origin. In this case, r2 can be found using the Pythagorean theorem since the charge is not located on the x-axis. The distance r2 can be calculated as:
r2 = sqrt((10 cm)^2 + (-5 cm)^2)
Once you have r2 calculated, you can plug the values into the formula to find E2.
Finally, to find the net electric field at the origin, you just need to add the electric fields E1 and E2 together. Since electric fields are vectors, make sure to pay attention to their direction and include the appropriate signs when adding them.
Now, as for the force exerted by each charge on the other, we can use Coulomb's law. The force between two charges is given by the formula:
F = k * |q1| * |q2| / r^2
where |q1| and |q2| are the magnitudes of the charges (without their signs), and r is the distance between them. In this case, the distance between the charges can be found using the Pythagorean theorem, just like before.
So, go ahead and plug in the values to calculate the forces each charge exerts on the other. And remember, a little humor always helps when dealing with electric charges!
To find the electric field due to the two charges at the origin, we need to calculate the individual electric fields produced by each charge and then sum them up.
Let's start by calculating the electric field produced by the 4.0 nC charge at the origin.
The formula to calculate the electric field due to a point charge is given by:
E = k * (q / r^2)
where:
E = electric field
k = Coulomb's constant (8.99 x 10^9 N m^2/C^2)
q = charge
r = distance between the charge and the point where the electric field is measured
In this case, q = 4.0 nC = 4.0 x 10^-9 C and the distance from the charge to the origin is 20 cm (or 0.20 m).
Plugging these values into the formula, we get:
E1 = (8.99 x 10^9 N m^2/C^2) * (4.0 x 10^-9 C) / (0.20 m)^2
Calculating this gives us:
E1 = 0.1798 N/C
Now let's calculate the electric field produced by the -3.0 nC charge at the origin.
Using the same formula, we have:
q = -3.0 nC = -3.0 x 10^-9 C
r = √[(10 cm)^2 + (-5 cm)^2] = √[100 cm^2 + 25 cm^2] = √[125 cm^2] = 11.18 cm = 0.1118 m
Plugging these values into the formula, we get:
E2 = (8.99 x 10^9 N m^2/C^2) * (-3.0 x 10^-9 C) / (0.1118 m)^2
Calculating this gives us:
E2 = -1.072 N/C
Since electric field is a vector quantity, we need to take into account the direction. In this case, the electric field produced by the positive charge is directed away from it, and the electric field produced by the negative charge is directed towards it.
Now we can find the net electric field at the origin by summing the individual electric fields:
E_net = E1 + E2
E_net = 0.1798 N/C + (-1.072 N/C)
E_net = -0.892 N/C
Therefore, the electric field due to these two charges at the origin is -0.892 N/C, directed towards the negative charge.
To find the force each charge exerts on the other, we can use Coulomb's law. The force between two charges is given by:
F = k * (|q1| * |q2|) / r^2
where:
F = force
k = Coulomb's constant (8.99 x 10^9 N m^2/C^2)
|q1|, |q2| = the magnitudes of the charges
r = distance between the charges
Let's calculate the force each charge exerts on the other.
For the 4.0 nC charge:
|q1| = 4.0 nC = 4.0 x 10^-9 C
|q2| = 3.0 nC = 3.0 x 10^-9 C
r = √[(20 cm)^2 + 5 cm)^2] = √[(400 cm^2 + 25 cm^2)] = √425 cm^2 = 20.62 cm = 0.2062 m
Plugging these values into the formula, we get:
F1 = (8.99 x 10^9 N m^2/C^2) * (4.0 x 10^-9 C) * (3.0 x 10^-9 C) / (0.2062 m)^2
Calculating this gives us:
F1 = 1.2036 x 10^-5 N
For the -3.0 nC charge:
|q1| = 4.0 nC = 4.0 x 10^-9 C
|q2| = 3.0 nC = 3.0 x 10^-9 C
r = √[(20 cm)^2 + (5 cm)^2] = √[(400 cm^2 + 25 cm^2)] = √425 cm^2 = 20.62 cm = 0.2062 m
Plugging these values into the formula, we get:
F2 = (8.99 x 10^9 N m^2/C^2) * (3.0 x 10^-9 C) * (4.0 x 10^-9 C) / (0.2062 m)^2
Calculating this gives us:
F2 = 1.2036 x 10^-5 N
Therefore, the force each charge exerts on the other is 1.2036 x 10^-5 N, directed towards each other.
To find the electric field due to these two charges at the origin, we need to calculate the electric field contribution from each charge separately and then add them together.
1. Electric Field due to the +4.0 nC charge:
The electric field due to a point charge is given by Coulomb's Law, which states that the electric field (E) at a certain point is equal to the magnitude of the charge (q) divided by the distance squared (r^2), multiplied by a constant (k), which is known as the Coulomb's constant.
Using this formula, we can calculate the electric field due to the +4.0 nC charge at the origin (0, 0):
E_1 = k * (q_1 / r_1^2),
where q_1 = +4.0 nC and r_1 = 0.20 meters (distance from the charge to the origin).
2. Electric Field due to the -3.0 nC charge:
Again, using Coulomb's Law, we can calculate the electric field due to the -3.0 nC charge at the origin (0, 0):
E_2 = k * (q_2 / r_2^2),
where q_2 = -3.0 nC and r_2 = sqrt((0.10 meters)^2 + (0.05 meters)^2) = 0.1118 meters (distance from the charge to the origin).
Next, we need to consider the direction of these electric fields. The +4.0 nC charge is positive, so the electric field points away from it (radially outward). The -3.0 nC charge is negative, so the electric field points toward it (radially inward).
Now, we can add the electric fields together to get the net electric field at the origin:
E_net = E_1 + E_2.
To find the force each charge exerts on the other, we can use Coulomb's Law again, but this time calculate the force (F) instead of the electric field.
Coulomb's Law states that the force between two charges (F) is proportional to the product of the charges (q_1 and q_2) divided by the square of the distance between them (r^2), multiplied by a constant (k).
Using this formula, we can calculate the force exerted by the +4.0 nC charge on the -3.0 nC charge, as well as the force exerted by the -3.0 nC charge on the +4.0 nC charge:
F_1 = k * (q_1 * q_2 / r_1^2),
F_2 = k * (q_1 * q_2 / r_2^2).
It's important to note that since forces are vector quantities, the direction of these forces depends on the charges (positive or negative) and the distances between them. The forces exerted on each charge are equal in magnitude but opposite in direction according to Newton's Third Law of Motion.
Please let me know if you need further assistance or any additional calculations!