A point charge Q1=-2µC is located at x=0, and a point charge Q2=-8µC is placed at on the x -axis of a cartesian coordinate system.The goal of this problem is to determine the electric field,E(x)=E(x)x^ , at various points along the -axis.

(a)What is E(x) (in N/C) for x -68.0 m ?

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What is E(x) (in N/C) at -0.42 m?

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What is E(x) (in N/C) at 65 m?

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(b) At what point (apart from,|x|=∞ )is ? Express your answer in meters.

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Please give answer.

b-0.5

To find the electric field, E(x), at various points along the x-axis, we can use the formula:

E(x) = k * (Q1 / r1^2 + Q2 / r2^2)

where:
- E(x) is the electric field at position x
- k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2)
- Q1 and Q2 are the charges at positions r1 and r2, respectively.

In this case, Q1 = -2 µC and Q2 = -8 µC.

(a) To find E(x) at x = -68.0 m:

Using the formula, we have:
E(x) = k * (Q1 / r1^2 + Q2 / r2^2)

Since Q1 is located at x = 0 and Q2 is located at x = -68.0 m, we have:
r1 = 0 and r2 = 68.0 m

Plugging the values into the formula, we get:
E(x) = (8.99 x 10^9 N m^2/C^2) * (-2 x 10^-6 C / (0^2) + (-8 x 10^-6 C) / (68.0^2 m^2))

E(x) = -2.12 x 10^5 N/C

Therefore, the electric field at x = -68.0 m is -2.12 x 10^5 N/C.

(b) To find the point where E(x) is zero (apart from |x| = ∞), we can set E(x) = 0 and solve for x.

Let's consider the second term in the bracket (Q2 / r2^2) since Q1 is located at x = 0:

0 = k * Q2 / r2^2

Solving for x, we have:
x = sqrt(k * Q2 / Q1)

Plugging in the values of k, Q2, and Q1, we get:
x = sqrt((8.99 x 10^9 N m^2/C^2) * (-8 x 10^-6 C / (-2 x 10^-6 C)))

x = sqrt(2.25) = 1.5 m

Therefore, apart from |x| = ∞, the point where E(x) is zero is at x = 1.5 m.

To solve this problem, we need to calculate the electric field at different points on the x-axis using the formula for electric field:

E(x) = k * (Q / r^2)

where E(x) is the electric field at point x, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge creating the electric field, and r is the distance between the point and the charge.

(a) Let's calculate the electric field at x = -68.0 m. Here, we have two point charges: Q1 = -2µC at x = 0 and Q2 = -8µC at x = 3 m.

First, we calculate the distance between the point and each charge:

r1 = |x - x1| = |-68.0 m - 0| = 68.0 m
r2 = |x - x2| = |-68.0 m - 3 m| = 71.0 m

Now we can plug these values into the formula:

E1 = k * (Q1 / r1^2) = 9 x 10^9 Nm^2/C^2 * (-2 x 10^-6 C) / (68.0 m)^2
E2 = k * (Q2 / r2^2) = 9 x 10^9 Nm^2/C^2 * (-8 x 10^-6 C) / (71.0 m)^2

Finally, we sum up the contributions from both charges to get the total electric field:

E(x) = E1 + E2

(b) To find the point on the x-axis where the electric field is zero, we need to equate the contributions from each charge:

E1 = E2

This can be expressed as:

k * (Q1 / r1^2) = k * (Q2 / r2^2)

Simplifying the equation, we get:

Q1 / r1^2 = Q2 / r2^2

By solving this equation, we can find the point where the electric field is zero.