The length of a vector arrow represents an magnitude
Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.
Explanation:
The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:
F = BqVSinØ
If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field
According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.
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Complete question:
if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.
Answer:
The mutual force between the two point charges is 319.64 N
Explanation:
Given;
distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m
value of the charges, q₁ and q₂ = 2 μC and - μ4 C
Apply Coulomb's law;
![F = \frac{k|q_1||q_2|}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bk%7Cq_1%7C%7Cq_2%7C%7D%7Br%5E2%7D)
where;
F is the force of attraction between the two charges
|q₁| and |q₂| are the magnitude of the two charges
r is the distance between the two charges
k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²
![F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bk%7Cq_1%7C%7Cq_2%7C%7D%7Br%5E2%7D%20%5C%5C%5C%5CF%20%3D%20%5Cfrac%7B8.99%2A10%5E9%20%2A4%2A10%5E%7B-6%7D%2A2%2A10%5E%7B-6%7D%7D%7B%281.5%2A10%5E%7B-2%7D%29%5E2%7D%20%5C%5C%5C%5CF%20%3D%20319.64%20%5C%20N)
Therefore, the mutual force between the two point charges is 319.64 N