Please help!
Two charges are placed on the x axis. One of the charges (q1 = +8.71C) is at x1 = +3.00 cm and the other (q2 = -28.6C) is at x2 = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.
Oh, I can help you with that! Let's tackle this electric field problem with a sprinkle of humor.
(a) At x = 0 cm, it's like the charges are getting ready for a grand party, but they have some serious differences! One charge is positive, and the other is negative. So, they attract each other but disagree on a lot of things, just like those neighbors who always argue about their lawn decorations.
The electric field due to q1 is directed towards the right (positive x-axis), while the electric field due to q2 is directed towards the left (negative x-axis). They're pulling in opposite directions, canceling each other out.
Therefore, the net electric field at x = 0 cm is 0N/C, which is as calm as a yoga session.
(b) Ah, at x = +6.00 cm, things get interesting! It's like the charges are playing a game of tug-of-war, but each charge wants to pull the rope in their own direction. The positive charge still attracts towards the right (positive x-axis), and the negative charge continues to attract towards the left (negative x-axis).
However, the electric field due to q1 is stronger since it's closer to x = +6.00 cm. So, it pulls a bit harder than q2, making the net electric field slightly to the right (positive x-axis).
But don't worry, it's not a strong push! It's like a gentle nudge from a friend trying to help you make a decision. So, the net electric field at x = +6.00 cm is a small positive value.
Hope that clears things up! Remember, even charges have their own electrifying humor.
To find the net electric field at a point, we need to calculate the electric field due to each individual charge and then add them vectorially.
Given:
Charge q1 = +8.71 C at x1 = +3.00 cm
Charge q2 = -28.6 C at x2 = +9.00 cm
To find the electric field at a point, we can use Coulomb's law:
Electric field due to a point charge q at a distance r is given by:
E = k * q / r^2
Where:
E is the electric field
k is the electrostatic constant (k = 9.0 x 10^9 Nm^2/C^2)
q is the charge
r is the distance from the charge to the point in question
Let's calculate the electric field due to q1 at x = 0 cm:
r1 = x1 - x = 3.00 cm - 0 cm = 3.00 cm = 0.03 m
E1 = k * q1 / r1^2
Substituting the given values:
E1 = (9.0 x 10^9 Nm^2/C^2) * (8.71 C) / (0.03 m)^2
Calculating this:
E1 = 9.0 x 10^9 Nm^2/C^2 * 8.71 C / (0.03 m)^2
E1 = [2.634 x 10^12 N/C]
Note: The direction of the electric field will be towards the positive x-axis since q1 is positive.
Next, let's calculate the electric field due to q2 at x = 0 cm:
r2 = x2 - x = 9.00 cm - 0 cm = 9.00 cm = 0.09 m
E2 = k * q2 / r2^2
Substituting the given values:
E2 = (9.0 x 10^9 Nm^2/C^2) * (-28.6 C) / (0.09 m)^2
Calculating this:
E2 = 9.0 x 10^9 Nm^2/C^2 * -28.6 C / (0.09 m)^2
E2 = [-1.540 x 10^12 N/C]
Note: The direction of the electric field will be towards the negative x-axis since q2 is negative.
To find the net electric field at x = 0 cm, we need to add the electric fields due to q1 and q2 vectorially.
Net Electric Field at x = 0 cm (E_net) = E1 + E2
Substituting the calculated values:
E_net = [2.634 x 10^12 N/C] + [-1.540 x 10^12 N/C]
Calculating this:
E_net = [1.094 x 10^12 N/C]
The net electric field at x = 0 cm is 1.094 x 10^12 N/C, directed towards the positive x-axis.
Now let's calculate the electric field at x = +6.00 cm:
For this calculation, we will use the same steps as before and substitute x = 6.00 cm.
Finding the electric field due to q1 at x = 6.00 cm:
r1 = x1 - x = 3.00 cm - 6.00 cm = -3.00 cm = -0.03 m
E1 = k * q1 / r1^2
Substituting the given values:
E1 = (9.0 x 10^9 Nm^2/C^2) * (8.71 C) / (-0.03 m)^2
Calculating this:
E1 = 9.0 x 10^9 Nm^2/C^2 * 8.71 C / (-0.03 m)^2
E1 = [-2.634 x 10^12 N/C]
Note: The direction of the electric field will be towards the negative x-axis since r1 is negative.
Finding the electric field due to q2 at x = 6.00 cm:
r2 = x2 - x = 9.00 cm - 6.00 cm = +3.00 cm = 0.03 m
E2 = k * q2 / r2^2
Substituting the given values:
E2 = (9.0 x 10^9 Nm^2/C^2) * (-28.6 C) / (0.03 m)^2
Calculating this:
E2 = 9.0 x 10^9 Nm^2/C^2 * -28.6 C / (0.03 m)^2
E2 = [-1.540 x 10^12 N/C]
Note: The direction of the electric field will be towards the negative x-axis since r2 is positive.
Now we can find the net electric field at x = 6.00 cm (E_net) by adding the electric fields due to q1 and q2 vectorially.
E_net = E1 + E2
Substituting the calculated values:
E_net = [-2.634 x 10^12 N/C] + [-1.540 x 10^12 N/C]
Calculating this:
E_net = [-4.174 x 10^12 N/C]
The net electric field at x = 6.00 cm is 4.174 x 10^12 N/C, directed towards the negative x-axis.
To find the net electric field at a certain point, you need to calculate the individual electric fields produced by each charge and then add them vectorially.
The formula for electric field due to a point charge is given by:
E = k * q / r^2
where E is the electric field, k is the electrostatic constant (9.0 x 10^9 N m^2/C^2), q is the charge, and r is the distance from the charge.
Let's calculate the electric field at x = 0 cm (point a):
For q1 at x1 = +3.00 cm:
r1 = x1 - x = 3.00 cm
E1 = k * q1 / r1^2
For q2 at x2 = +9.00 cm:
r2 = x2 - x = 9.00 cm
E2 = k * q2 / r2^2
To calculate the net electric field, we need to take into account the direction and add the magnitudes appropriately.
For point a, x = 0 cm:
E1 = k * q1 / r1^2 = (9.0 x 10^9 N m^2/C^2) * (8.71 C) / (0.03 m)^2
E2 = k * q2 / r2^2 = (9.0 x 10^9 N m^2/C^2) * (-28.6 C) / (0.09 m)^2
To find the net electric field, we add the electric fields due to q1 and q2:
E_net = E1 + E2
You can now substitute the values into the equations and calculate the net electric field at x = 0 cm.
E = k * q/r^2
The distances all have to be converted to meters because you want everything in SI units
a)
at x = 0, the field from the first charge is k * 8.71 / 0.03m^2
the field from the 2nd charge is k * -28.6 / 0.09m^2
So the total electric field is the sum of these numbers
k * 8.71 / 0.03^2 + -28.6 / 0.09^2
b) at x = 6.00 cm, the first charge is a distance of (6-3) = 3 cm away = 0.03 m, and the 2nd charge is at a distance
(9-6) = 3 cm = 0.03 m
So the total electric field is
k * 8.71 / 0.03^2 + k * -28.6 / 0.03^2