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A +18 nC charge is placed on the x-axis at x = 1.6 m, and a -18 nC charge is placed at x = -4.7 m. What is the magnitude of the electric field at the origin? Give your answer to one decimal place.

User WaeCo
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To find the magnitude of the electric field at the origin due to the two charges, we can use the formula for the electric field created by a point charge:

Electric Field (E) = k * |q| / r^2

Where:

k is Coulomb's constant, approximately 8.99 x 10^9 N m^2/C^2

|q| is the magnitude of the charge in Coulombs

r is the distance from the charge to the point where we want to find the electric field

Let's calculate the electric field at the origin due to the +18 nC charge at x = 1.6 m:

Charge |q| = 18 nC = 18 x 10^(-9) C

Distance r1 = 1.6 m

Electric Field (E1) = (8.99 x 10^9 N m^2/C^2) * (18 x 10^(-9) C) / (1.6 m)^2

Now, let's calculate the electric field at the origin due to the -18 nC charge at x = -4.7 m:

Charge |q| = 18 nC = 18 x 10^(-9) C

Distance r2 = 4.7 m

Electric Field (E2) = (8.99 x 10^9 N m^2/C^2) * (18 x 10^(-9) C) / (4.7 m)^2

Now, since the electric field is a vector quantity, we need to consider both the magnitudes and directions of the electric fields created by the +18 nC and -18 nC charges.

The electric field at the origin (E_total) is the vector sum of E1 and E2:

E_total = E1 + E2

Now, calculate the magnitude of E_total:

|E_total| = √(E1^2 + E2^2)

Substitute the calculated values and compute |E_total| to one decimal place:

|E_total| ≈ 3.2 x 10^6 N/C

The magnitude of the electric field at the origin is approximately 3.2 x 10^6 N/C.

User BillyJean
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