Please show all work clearly, thank you! A point charge of +3e is at the origin

Question

Please show all work clearly, thank you!

A point charge of +3e is at the origin and a second point charge of-2e is on the x-axis at x = a. (a) derive the potential function V(x) versus x for all points on the x axis. (b) At what point or points, if any, is V = zero on the x axis? (c) What point or points, if any, on the x-axis is the electric field zero? Are these locations the same locations found in Part (b)? Explain your answer. (d) How much work is needed to bring a third charge +e to the point x = 1/2 a on the x-axis?

Explanation / Answer

For the two charges, r = x a and x respectively and the electric potential at x is the algebraic sum of the potentials at that point due to the charges at x = a and x = 0. We can use the graph and the function found in Part (a) to identify the points at which V(x) = 0. We can find the work needed to bring a third charge +e to the point x a2 1 = on the x axis from the change in the potential energy of this third charge.

(a) The potential at x is the sum of the potentials due to the point charges +3e and 2e:

V x =k(3e)/|x|+k(-2e)/|x-a|

The following graph of V(x) for ke = 1 and a =1 was plotted using a spreadsheet program.

(b) From the graph we can see that V(x) = 0 when: x = ±

Examining the function, we see that V(x) is also zero provided:

3/|x|-2/|x-a|=0

For x > 0, V(x) = 0 when: x = 3a

For 0 < x < a, V(x) = 0 when: x = 0.6a

(c) The electric field at x is the sum of the electric fields due to the point charges +3e and 2e:

E9x)=k(3e)/x^2+k(-2e)/|x-r|^2

Setting E(x) = 0 and simplifying yields:

x^2 6ax + 3a^2 =0

Solve this equation to find the points on the x-axis where the electric field is zero: x = 5.4a and x = 0.55a

Note that the zeros of the electric field are different from the zeros of the electric potential. This is generally the case although, in special cases, they can be the same.

(d) Express the work that must be done in terms of the change in potential energy of the charge:

W=delta U=qV(1/2a)

Evaluate the potential at x=1/2a

V(1/2a)=k(3e)/|1/2a|+k(-2e)/|1/2a-a|

=2ke/a

Substitute to obtain:

W=e(2ke)/a=2ke^2/a

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.