The nucleus of a copper atom contains 29 protons and has a radius of 4.8 x 10^-15 m. how much work (in electron volts) is done by the electric force as a proton is brought from infinity, where it is at rest, to the "surface" of a copper nucleus?

The copper nucleus and the proton we're bringing in will repel each other. The electric force will point in the opposite direction to us moving the proton toward the nucleus, so a net positive amount of external work will have to be done against the electric field. Thus the electric field does a net negative amount of work.

The work we have to do against the field is given by:

W = ΔVq

W = work, ΔV = potential difference, q = proton charge

If the 0 potential is defined to be infinitely far away, then the potential difference ΔV is given by:

ΔV = kQ/r

Q = nucleus charge, r = radius of atom

Make a substitution:

W = kQq/r

Given values:

k = Coulomb constant, Q = 29e (elementary charges), q = 1e, r = 4.8×10⁻¹⁵m

Plug in and solve for W:

W = (9×10⁹)(29)(1)/(4.8×10⁻¹⁵)

W = 8.7MeV

Put a negative sign in front of W, and now we have the work done by the electric field, -8.7MeV

good looking response

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg