Complete.
1 answer:
Answer:
For a positive point charge, the lines radiate outward. While, for a negative point charge, the lines converge inward.
Explanation:
The electric field produced by a single point charge has a radial shape and follows an inverse square law. The magnitude is given by
where
k is the Coulomb's constant
q is the magnitude of the charge
r is the distance from the charge
The direction of the field depends on the sign of the charge. In particular, we have:
- For a positive charge, the direction of the field is out of the charge
- FOr a negative charge, the direction of the field is towards the charge
This is due to the fact that the direction of the field shows the direction of the force that a positive test charge would experience in this field. Therefore, since a positive test charge would feel a repulsive force in the field produced by a positive charge, then the direction of the field is outward; on the other hand, a positive test charge would feel an attractive force in the field produced by a positive charge, therefore the direction of the field is inward.
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Answer:
electricfield at (0,0,0) is Et = 2 k q / a²
Explanation:
For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity
For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation
E = k q / r²
Point (0,0,0)
We calculate for the charge -q which is at a distance R = a
E1 = k (-q) / a²
E1 = - kq / a²
As the test charge is positive in the field it goes to the left, attractive force
We calculate for the charge that is also at R = a
E2 = k q / a²
This field goes to the left, repulsive force
We find the total electric field
Et = E1 + E2
Et = kq / a² + kq / a²
Et = 2 k q / a²
Point (0,0, R)
We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a
E1 = k q / (R + a)²
E2 = kq / (R-a)²
Et = kq [1 / (R + a)² + 1 / (R-a)²]
Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}
Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}
If we use the condition that R> a we can despise in the patents "a"
(R² + a²) = R² (1+ a² / R²) ≈ R²
(R + a)² = R² (1 + a / R)² ≈ R²
(R- a)² = R² (1-a / R)² ≈ R²
Substituting in the total electric field
Et = kq {2 R²) / [R²R²]}
Et =kq 2 / R²
The net force on the barge is 8000 N
Explanation:
In order to find the net force on the badge, we have to use the rules of vector addition, since force is a vector quantity.
In this problem, we have two forces:
- The force of tugboat A,
, acting in a certain direction
- The force of tugboat B,
, also acting in the same direction
Since the two forces act in the same direction, this means that we can simply add their magnitudes to find the net combined force on the barge. Therefore, we get
and the direction is the same as the direction of the two forces.
Learn more about forces:
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