A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 5600 N/C and is directed in the positive x direction. A point charge -9.6 × 10-9 C is placed at the origin. Find the magnitude of the net electric field at (a) x = -0.20 m, (b) x = +0.20 m, and (c) y = +0.20 m.
Well, well, well...it seems we have an electric field and a point charge playing hide and seek! Let's see if we can find them, shall we?
(a) To find the magnitude of the net electric field at x = -0.20 m, we'll need to use a bit of math magic. We know the magnitude of the electric field is 5600 N/C, and since it's directed in the positive x direction, we can't let it go unnoticed. Now, the point charge is at the origin, which means it's right in the middle of this field's playground.
Since the electric field is uniform, its magnitude remains the same no matter where we are in the x, y plane. So, at x = -0.20 m, the magnitude of the net electric field will still be 5600 N/C, ready to make anyone's hair stand on end!
(b) Now, let's move to x = +0.20 m. Being the same magnitude electric field, it can't really change its nature that easily. So, at x = +0.20 m, guess what?! You got it, the magnitude of the net electric field will still be 5600 N/C. You can't escape the uniformity, my friend.
(c) Finally, let's take a stroll down y = +0.20 m to find out what happens there. Now, keep in mind, we're still dealing with a uniform electric field in the x, y plane. So, no matter how far you wander in the y direction, it won't affect the magnitude of the electric field. The magnitude will remain a steadfast 5600 N/C, undeterred by your yearnings for something different.
So, there you have it! At x = -0.20 m, x = +0.20 m, and y = +0.20 m, the magnitude of the net electric field will all be 5600 N/C, keeping up with its uniformity. Now, go forth and spread the electrifying news!
To find the magnitude of the net electric field at different points in the x-y plane, we need to consider the contribution of the uniform electric field and the point charge.
The electric field due to a uniform electric field can be found using the equation:
E_uniform = magnitude of electric field × unit vector in the direction of the field
In this case, the magnitude of the uniform electric field is 5600 N/C, and since it is directed in the positive x direction, the unit vector in the direction of the field is ⟨1, 0⟩.
The electric field due to a point charge can be found using Coulomb's Law:
E_point = k × (charge / distance^2) × unit vector in the direction from the point charge
Where:
- k is the Coulomb's constant (k ≈ 9 × 10^9 N m^2 / C^2)
- charge is the magnitude of the point charge (-9.6 × 10^-9 C)
- distance is the distance between the point charge and the point where we want to find the electric field
- unit vector is the normalized vector pointing from the charge to the point where we want to find the electric field
Now, let's calculate the magnitude of the net electric field at the given points.
(a) At x = -0.20 m:
The electric field due to the uniform field will be:
E_uniform = 5600 N/C × ⟨1, 0⟩ = ⟨5600, 0⟩ N/C
The distance between the point charge at the origin and the point (x = -0.20 m, y = 0) is 0.20 m. The unit vector in the direction from the charge can be calculated as:
unit vector = (-0.20 m - 0 m, 0 - 0 m) / 0.20 m = ⟨-1, 0⟩
The electric field due to the point charge will be:
E_point = k × (9.6 × 10^-9 C / 0.20 m^2) × ⟨-1, 0⟩
The net electric field at x = -0.20 m will be the vector sum of the electric fields due to the uniform field and the point charge:
E_net = E_uniform + E_point
Calculate the magnitude of E_net to get the answer.
(b) At x = +0.20 m:
Follow the same steps as in part (a), but with the unit vector in the direction from the charge being ⟨1, 0⟩.
(c) At y = +0.20 m:
The electric field due to the uniform field will be:
E_uniform = 5600 N/C × ⟨1, 0⟩ = ⟨5600, 0⟩ N/C (same as in part (a))
The distance between the point charge at the origin and the point (x = 0, y = +0.20 m) is 0.20 m. The unit vector in the direction from the charge can be calculated as:
unit vector = (0 m - 0 m, +0.20 m - 0 m) / 0.20 m = ⟨0, 1⟩
The electric field due to the point charge will be:
E_point = k × (9.6 × 10^-9 C / 0.20 m^2) × ⟨0, 1⟩
The net electric field at y = +0.20 m will be the vector sum of the electric fields due to the uniform field and the point charge:
E_net = E_uniform + E_point
Calculate the magnitude of E_net to get the answer.