Four equal negative point charges are located at the corners of a square, their positions in the xy-plane being (1,1), (−1,1), (
−1,−1), (1,−1).
The electric field on the x-axis at (1,0) points in the same direction as:
A) -j
B) i
C) -i
D) j
The electric field on the x-axis at (1,0) points in the same direction as:
A) -j
B) i
C) -i
D) j
2 answers:
Answer:
Result is C, -i^
Explanation:
The electric field is a vector quantity given by
E = k ∑
Where k is the Coulomb constant that is worth 8,988 10⁹ N / m² C², q is the charge and r is the distance from the charge to the point of interest
Let's apply this formula to our case
E = E1 + E2 + E3 + E4
the bold are vectors
The point where we want the field is r₀ = 1 i ^ + 0 j ^. Let's look for the distances
r² = (x-x₀)² + (y-y₀)²
r₁² = (1-1)² + (1-0)²
r₂² = (-1-1)² + (1-0)²
r₃² = (-1-1)² + (-1-0)²
r₄² = (1-1)² + (-1-0)²
r₁² = 1
r₂² = 5
r₃² = 5
r₄² = 1
The charge proof is always considered positive
Let's analyze the direction of each field
²E1 between the charge in position 1 and the point of interest the force is attractive, charge with different sign directed on the y-axis (vertical), whereby E₁ = E₁ j ^
E4 between charge 4 and the test charge, incoming field, directed on the vertical axis and E₄ = -E₄ j ^
E2 between the charge in 3 and the test charge in r, the field is incoming, with the direction on the line between 3 and 1, let's look with trigonometry
tan θ₂ = y / x
θ₂ = tan⁻¹ y / x
θ₂ = tan⁻¹ 1 / 2
θ₂ = 26.57º
E3 between charge 3 and test carge in ro, incoming field, with address
tan θ₃ = y / x
θ₃ = tan⁻¹ y / x
θ₃ = tan⁻¹ (-1/2)
θ₃ = -26.57º
From the symmetry and the distances we see that the field E₁ and E₄ are of the same magnitude and opposite direction whereby their sum is zero.
Fields E2 and E3 have components x and y, since the distances are equal the magnitude of the two fields is equal. The sum of the components in y is zero and the components in x
E_total = 2 E₂ cos 26.57
Total_ = 2 k q / r₂² cos 26.57 (- i ^)
Total_ = 2 8,988 10⁹ q / 5 cos 26.57 (- i ^)
Total_ = 16.08 10⁹ q (-i ^)
The address of this total field is –i ^
Result is C
Answer: c) -i
Explanation:
Since the particles carry the same charges they will all repel each other,
Since the particles are at equal distance from each other, the magnitude of their repelling forces on each other will be equal assuming they are all of equal mass.
Each perticle will move away at a 90° direction from the center and this puts the electric field at the center to sum up to zero.
All field will point towards the point of least potential hence the electric field at point (1, - 1) will point towards -I direction which is along the x-axis towards the zero point.
Mathematically, force on each particle Fp = -Qq/r^2 (coulumb's law)
Assuming each has a charge of 1C
The force field between point (1, 0) and point (-1, 1) or (-1, - 1) will vary as inverse square of their distance,
Since they are 5 units apart (length of their diagonal), they experience a repulsion of the magnitude 0.04 (1/25)
For points (1, 0) and (1, 1) or (1, - 1) the magnitude is 1 since they are 1 unit apart. This is 25 times stronger than for the above.
This means the field will move towards the lesser field which is toward -i
See image below
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f = 1/T [1/s] or [s^-1] or [Hz]
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ii) replacing in the derivated equation.
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2
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The mosquitoes that can sit on the water.
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Answer:
He should run at least at 1.5 m/s
The diver will enter the water at an angle of 87° below the horizontal.
Explanation:
Hi there!
The position and velocity of the diver are given by the following vectors:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v = (v0x, v0y + g · t)
Where:
r = position vector at time t
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
v = velocity vector at time t
Please, see the attached figure for a description of the problem. Notice that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0.
We know that, to clear the rocks, the position vector r final (see figure) should be:
r final = ( > 5.0 m, -50 m)
So let´s find first at which time the y-component of the vector r final is - 50 m:
y = y0 + v0y · t + 1/2 · g · t²
-50 m = 2.1 m/s · t - 1/2 · 9.8 m/s² · t²
0 = -4.9 m/s² · t² + 2.1 m/s · t + 50 m
Solving the quadratic equation
t = 3.4 s
Now, we can calculate the initial horizontal velocity using the equation of the x-component of the position vector knowing that at t =3.4 the horizontal component should be greater than 5.0 m:
x = x0 + v0x · t (x0 = 0)
5.0 m < v0x · 3.4 s
v0x > 5.0 m / 3.4 s
v0x > 1.5 m/s
The initial horizontal velocity should be greater than 1.5 m/s
To find the angle at which the diver enters the water, we have to find the magnitude of the final velocity (vector vf in the figure). We already know the magnitude of the x-component of the vector vf, since the horizontal velocity is constant. So:
vfx > 1.5 m/s
Now, let´s calculate vfy:
vfy = v0y + g · t
vfy = 2.1 m/s - 9.8 m/s² · 3.4 s
vfy = -31 m/s
Let´s calculate the minimum magnitude that the final velocity will have if the diver safely clears the rocks. Let´s consider the smallest value allowed for vfx: 1.5 m/s. Then:
Then the final velocity of the diver will be greater or equal than 31 m/s.
To find the angle, we have to use trigonometry. Notice in the figure that the vectors vf, vfx and vy form a right triangle in which vf is the hypotenuse, vfx is the adjacent side and vfy is the opposite side to the angle. Then:
cos θ = adjacent / hypotenuse = vfx / vf = 1.5 m/s / 31 m/s
θ = 87°
The diver will enter the water at an angle of 87° below the horizontal.
