At what point(s) along the x axis is the potential zero? Determine the x-coordin

Question

At what point(s) along the x axis is the potential zero? Determine the x-coordinate(s) of the point(s).

At what point(s) along the x axis is the electric field zero? Determine the x-coordinate(s) of the point(s) Express your answer using two significant figures. If there is more than one answer, enter your answers in ascending order separated by commas. At what point(s) along the x axis is the potential zero? Determine the x-coordinate(s) of the point(s).Express your answer using two significant figures. If there is more than one answer, enter your answers in ascending order separated by commas.

Explanation / Answer

part A:

as field due to positive charge is away from the charge and negative charge is towards the charge,

total field will be zero either to left of Q1 or to the right of Q2.

but as field magnitude is inversely proportional to square of the distance,

and Q1 magnitude is higher than magnitude of Q2

distance from Q1 should be more than distance from Q2

hence the point should be to the right of Q2.

let distance from Q2 be x.

then equating field magnitude due to both the charges:

k*Q1/(3+x)^2=k*Q2/x^2

==>3/(3+x)^2=1.4/x^2

==>3*x^2=1.4*(9+x^2+6*x)

==>1.6*x^2-8.4*x-12.6=0

hence x=6.4676 cm

part B:

let at x=d potential is 0.

where d>0

then writing total potential and equating to zero,

(k*Q1/(3+d))+(k*Q2/d)=0

==>3/(3+d)=1.4/d

==>3*d=1.4*(3+d)

==>3*d=4.2+1.4*d

==>d=2.625

hence at x=2.625 cm, potential is 0.

there is also a possibility of potential being zero in between Q1 and Q2.

let distance of that point from Q2 be d.

then distance from Q1=3-d

hence k*Q1/(3-d) + k*Q2/d=0

==>3/(3-d)=1.4/d

==>3*d=1.4*(3-d)

==>4.4*d=1.4*3

==>d=0.9545 cm

hence at x=-0.9545 cm potential is zero.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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