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Two charges, one of 2C at the origin and another of 8Clocatedatx=10 m, are held in place. What is the electric field at a point x= −5 m ? 0.04kN/C 0.12kN/C 0.33kN/C 0.40kN/C 0.93kN/C

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Answer:

The closest option to the calculated electric field is 0.40 kN/C.

Step-by-step explanation:

To calculate the electric field at a point, we can use Coulomb's Law, which states that the electric field due to a point charge is given by:

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

where E is the electric field, k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the point charge.

In this case, we have two charges:

Q1 = 2C at the origin (x = 0 m),

Q2 = 8C at x = 10 m,

We want to find the electric field at x = -5 m.

First, let's calculate the electric field due to Q1 at x = -5 m:

r1 = |-5 - 0| = 5 m (distance from Q1 to the point)

E1 = k * (Q1 / r1^2)

Substituting the values:

E1 = (8.99 x 10^9 Nm^2/C^2) * (2C / (5m)^2)

E1 = 8.99 x 10^9 Nm^2/C^2 * 2C / 25m^2

E1 ≈ 1.4392 x 10^8 N/C

Next, let's calculate the electric field due to Q2 at x = -5 m:

r2 = |-5 - 10| = 15 m (distance from Q2 to the point)

E2 = k * (Q2 / r2^2)

Substituting the values:

E2 = (8.99 x 10^9 Nm^2/C^2) * (8C / (15m)^2)

E2 = 8.99 x 10^9 Nm^2/C^2 * 8C / 225m^2

E2 ≈ 3.1956 x 10^7 N/C

The total electric field at x = -5 m is the sum of the electric fields due to Q1 and Q2:

E_total = E1 + E2

E_total ≈ 1.4392 x 10^8 N/C + 3.1956 x 10^7 N/C

E_total ≈ 1.7588 x 10^8 N/C

Rounding to two significant figures, the electric field at x = -5 m is approximately 1.8 x 10^8 N/C, which is equivalent to 0.18 kN/C.

Therefore, the closest option to the calculated electric field is 0.40 kN/C.

User Dion V
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