How do you find the direction and magnitude from a nucleus?

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SUMMARY

The discussion focuses on calculating the electric field and force exerted by a nucleus with a charge of 65 protons. The electric field (E) at a distance of 4.9 x 10^-10 m from the nucleus is calculated using the formula E = KQ/d², where K is 9 x 10^9 N m²/C². The calculated electric field magnitude is 2.44 x 10^30 N/C. Additionally, the force on an electron at this distance is derived from the electric field, emphasizing the need to understand the charge of the protons in coulombs.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with Coulomb's law
  • Basic knowledge of atomic structure and charge
  • Proficiency in unit conversions, specifically from protons to coulombs
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  • Study the concept of electric field strength and its calculation
  • Explore the relationship between electric fields and forces on charged particles
  • Investigate the conversion of atomic charge (protons) to coulombs
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Students studying physics, particularly those focusing on electromagnetism, as well as educators seeking to clarify concepts related to electric fields and forces in atomic structures.

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Homework Statement



The nucleus of an atom has a charge of 65 protons.

(a) What are the direction and magnitude of the electric field at 4.9 x 10^-10 m from the nucleus?

(b) What are the direction and magnitude of the force exerted on an electron located at this distance?


Homework Equations


E= KQ/d^2
K= 9 x 10^9


The Attempt at a Solution



(a) (9 x 10^9)(65)/(4.9 x 10^-10)^2
= 2.44e30 N

Please help clarify if (a) is correct, and what exactly is (b) asking, if any different from (a)?
 
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I think you need to understand what the charge of 65 protons is.
http://en.wikipedia.org/wiki/Proton

It is not 65 coulombs that are the units of the equation you used.

Having found the field, then b) is asking you to find the force.
 

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