1. The problem statement, all variables and given/known data The Nucleus of a helium ion contains two protons approximately 5x10-15m apart. They can be considered to be surrounded by one orbiting electron approximately 5x10-11m away. Calculate the following: a) The electrostatic force between the two protons b) The electrostatic force on the electron caused by the nucleus c) The electrical potential energy of the electron 2. Relevant equations Equations & Values F=kQq/r2 E=kQ/r2 V=kQ/r e=-1.6x10-19C k=8.99x109Nm2C-2 r (between protons)=5x10-15m r (electron)=5x10-11m 3. The attempt at a solution I am confident at my attempt at the answer for part a F=(8.99x109)(-1.6x10-19)(-1.6x10-19)/(5x10-15)2 F=9.2N (repulsive) For part b, I am slightly unsure as to what I am doing as I don't particularly understand what is it asking me. As the nucleus has the two protons within it, does this mean that it has double the charge? e.g. e(nucleus) = 2(-1.6x10-19)C = -3.2x10-19C so F=(8.99x109)(-3.2x10-19)(-1.6x10-19)/(5x10-11)2 F=1.8x10-7N (repulsive) Or am i missing something I should be doing to calculate the charge of the nucleus? For part C, my immediate thought is that: V=(8.99x109)(-1.6x10-19)/5x10-11 V=-2.9x10-7V (attractive) But, given that this answer is worth 3 marks, I am thinking this answer is too simple and I must be missing something here. I am sorry for this simplistic nature of this question, but in all honesty we have not been told anything on this subject and I am trying to fill in the blanks left by our tutors. Thank you in advance.