The minimum aperture diameter so that the cells can be resolved.

Question

An electron has an initial speed of 5.26 x 100 m/s in a uniform 5.73 x 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? opposite direction to the electron's initial velocity same direction as the electron's initial velocity not enough information to decide × What is the direction of the force on the electron? How does it compare to the direction of the electric field, considering the sign of the electron's charge? (b) How far does the electron travel before coming to rest? 0.0781 × What kinematic equation is relevant here? How do you calculate the force due to the electric field? m (c) How long does it take the electron to come to rest? 5.27e8 What is the final velocity of the electron? s (d) What is the electron's speed when it returns to its starting point? 5.26e6 m/s

Answer 1

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Chapter 36, Problem 11PQ

To determine

The minimum aperture diameter so that the cells can be resolved.

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An electron has an initial speed of 5.26 x 100 m/s in a uniform 5.73 x 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? opposite direction to the electron's initial velocity same direction as the electron's initial velocity not enough information to decide × What is the direction of the force on the electron? How does it compare to the direction of the electric field, considering the sign of the electron's charge? (b) How far does the electron travel before coming to rest? 0.0781 × What kinematic equation is relevant here? How do you calculate the force due to the electric field? m (c) How long does it take the electron to come to rest? 5.27e8 What is the final velocity of the electron? s (d) What is the electron's speed when it returns to its starting point? 5.26e6 m/s

(a) What magnitude point charge creates a 90,000 N/C electric field at a distance of 0.235 m? 5.53e-7 C (b) How large is the field at 22.2 m? 9e4 Using the equation for the electric field due to a point charge, and knowing the charge from part (a), can you solve for the field? N/C

No chatgpt pls will upvote Already got wrong chatgpt answer