A +5.00 C charge is located at the origin. A-3.00C charge is placed at x = 1.00 m. At what finite distance(s) along the x-axis will the electric field be equal to zero?

Question

in simple steps

Answer 1

To find the distance along the x-axis where the electric field is equal to zero, we can use the equation for electric field due to a point charge:

E = k * Q / r^2

where E is the electric field, k is the Coulomb's constant (9.0 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the charge.

Given:
Q1 = +5.00 C (charge at the origin)
Q2 = -3.00 C (charge at x = 1.00 m)

To find the distance where the electric field is zero, we need to find the point on the x-axis where the electric field due to Q1 and Q2 cancel out.

Let's consider the electric field due to Q1 at a distance x from the origin:

E1 = k * Q1 / x^2

Similarly, let's consider the electric field due to Q2 at a distance (1.00 - x) from the origin:

E2 = k * Q2 / (1.00 - x)^2

For the total electric field to be zero, the magnitudes of E1 and E2 must be equal. Therefore, we can set up the equation:

E1 = E2

k * Q1 / x^2 = k * Q2 / (1.00 - x)^2

Canceling out the k in both sides of the equation:

Q1 / x^2 = Q2 / (1.00 - x)^2

Now, we can substitute the values:

+5.00 / x^2 = -3.00 / (1.00 - x)^2

Cross-multiplying:

+5.00 * (1.00 - x)^2 = -3.00 * x^2

Expanding and rearranging the equation:

5.00 - 10.00x + 5.00x^2 = -3.00x^2

8.00x^2 - 10.00x + 5.00 = 0

This equation is a quadratic equation. To solve it, we can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 8.00, b = -10.00, and c = 5.00

Plugging in these values:

x = (-(-10.00) ± √((-10.00)^2 - 4(8.00)(5.00))) / (2(8.00))
x = (10.00 ± √(100.00 - 160.00)) / 16.00
x = (10.00 ± √(-60.00)) / 16.00

Since the square root of a negative number is not a real number, this equation has no real solutions. Therefore, there are no finite distances along the x-axis where the electric field is zero in this scenario.