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BlackZzzverrR [31]
3 years ago
12

At a given instant, the force on an electron is in the +z-direction (out of the page), which the electron is moving in the +x-di

rection (to the right). What can we say about the B field responsible for this force on the electron? (The y-direction is up on the page.)
Physics
1 answer:
Hitman42 [59]3 years ago
3 0

Answer:

The direction of the B-field is in the +y-direction.

Explanation:

The corresponding formula is

F_B = qv\times B

This means, we should use right-hand rule.

Our index finger is pointed towards +x-direction (direction of velocity),

our middle finger should point towards the direction of the B-field,

and our thumb should point towards the +z-direction (direction of the force).

Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.

\^{x} \times \^{y} = \^{z}

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Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
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Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

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Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

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The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

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