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Delvig [45]
3 years ago
12

Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has t

he largest force exerted on it by the magnetic field
Physics
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

case 1 of physics is the answer

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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
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Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

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