A charge +q is at the origin. A charge -2q is at x = 4.80 m on the +x axis.
(a) For what finite value of x is the electric field zero?
(b) For what finite values of x is the electric potential zero? (Note: Assume a reference level of potential V = 0 at r = ∞.)
Smallest value of x:
Largest value of x:
To find the value of x where the electric field is zero, we need to consider the electric fields produced by the charges +q and -2q at any point on the +x axis. The electric field due to a point charge q is given by the formula:
E = k * (q / r²),
where k is the Coulomb's constant (k = 9 × 10^9 N m²/C²) and r is the distance between the charge and the point where the electric field is being measured.
(a) Let's consider a point P on the +x axis, at a distance x from the origin. The electric field due to the charge +q at point P is given by:
E₁ = k * (q / r₁²),
where r₁ = x.
The electric field due to the charge -2q at point P is given by:
E₂ = k * (-2q / r₂²),
where r₂ = (4.80 - x) (distance from the charge -2q to point P).
Since the electric field at point P is the sum of the electric fields due to the individual charges, we can write:
E = E₁ + E₂.
0 = k * (q / x²) + k * (-2q / (4.80 - x)²).
To find the value of x where E = 0, we need to solve this equation. Multiply through by x² (to eliminate the variables in the denominator) and simplify:
0 = q * (4.80 - x)² - 2q * x².
Expanding (4.80 - x)² gives:
0 = q * (23.04 - 9.6x + x²) - 2q * x².
Distribute q through the parentheses:
0 = 23.04q - 9.6q * x + q * x² - 2q * x².
Combine like terms:
0 = (q - 2q) * x² + (-9.6q) * x + 23.04q.
Simplify further:
0 = -q * x² - 9.6q * x + 23.04q.
Divide through by q (since q ≠ 0):
0 = -x² - 9.6x + 23.04.
Now we have a quadratic equation. To solve it, we can factor, use the quadratic formula, or approximate using numerical methods. In this case, let's use the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a).
For our equation, the coefficients are:
a = -1,
b = -9.6, and
c = 23.04.
Substituting these values into the quadratic formula:
x = (-(-9.6) ± √((-9.6)² - 4(-1)(23.04))) / (2(-1)).
Simplifying:
x = (9.6 ± √(92.16 + 92.16)) / (-2).
x = (9.6 ± √(184.32)) / (-2).
x = (9.6 ± 13.57) / (-2).
There are two possible solutions:
x₁ = (9.6 + 13.57) / (-2) = 11.085 (rounded to three decimal places),
x₂ = (9.6 - 13.57) / (-2) = -1.985 (rounded to three decimal places).
Since we are looking for a finite value of x on the +x axis, the solution is:
(a) The electric field is zero at x = 11.085 m.
To find the value of x where the electric potential is zero, we can use a similar approach. The electric potential at a point due to a point charge q is given by the formula:
V = k * (q / r),
where r is the distance between the charge and the point where the potential is being measured.
(b) For the electric potential to be zero, the net potential from the charges +q and -2q should cancel each other. Since the potential is a scalar quantity, we just need to consider the magnitudes.
The potential due to the charge +q at point P is given by:
V₁ = k * (q / r₁),
where r₁ = x.
The potential due to the charge -2q at point P is given by:
V₂ = k * (-2q / r₂),
where r₂ = (4.80 - x).
The total potential at point P is the sum of the potentials due to the individual charges:
V = V₁ + V₂.
0 = k * (q / x) + k * (-2q / (4.80 - x)).
To find the value(s) of x where V = 0, we can solve this equation. Multiply through by x (to eliminate the variables in the denominators) and simplify:
0 = k * q * (4.80 - x) - 2k * q * x.
Divide through by kq (since kq ≠ 0):
0 = (4.80 - x) - 2x.
Simplify further:
0 = 4.80 - x - 2x.
Combine like terms:
0 = 4.80 - 3x.
Rearrange the equation:
3x = 4.80.
Divide through by 3:
x = 4.80 / 3.
x = 1.60.
Therefore, the electric potential is zero at x = 1.60 m.
To find the smallest and largest values of x where the electric potential is zero, we need to consider the range of x where V = 0. Since the charges are located at the origin and x = 4.80 m, the smallest and largest values of x are:
Smallest value of x = 1.60 m.
Largest value of x = 4.80 m.