Question

A 28-cm long charged wire with a total charge of 84 nC is bent to form three quarters of a circle and placed symmetrically with respect to the X-axis.

Assuming that the charge is uniformly distributed along the length of the wire, what is the magnitude of the electric field in kV/m at the center of this circle? Provide your answer accurately to two decimal places.

Answer 1

The magnitude of the electric field in kV/m at the center of the circle is approximately 4.78.

Step-by-step explanation:

To find the magnitude of the electric field at the center of the given circle, we can use the concept of the electric field due to a charged wire.

First, let's calculate the charge per unit length of the wire. We are given that the electric field at 2 cm from the center of a similar wire is 3 N/C. Using this information, we can use the following equation:

E = λ / (2πε₀r)

Where E is the electric field, λ is the charge per unit length, ε₀ is the permittivity of free space, and r is the distance from the wire.

Plugging in the values, we can solve for λ:

3 N/C = λ / (2πε₀(0.02 m))

λ ≈ (3 N/C)(2πε₀(0.02 m)) ≈ 2.39 × 10-5 C/m

Now, to find the magnitude of the electric field at the center of the given circle, we can use the following equation:

E = λ / (4πε₀(0.14 m))

Plugging in the values, we can solve for E:

E ≈ (2.39 × 10-5 C/m) / (4πε₀(0.14 m)) ≈ 4.78 × 103 N/C

Converting this value to kV/m, we divide by 1000:

E ≈ 4.78 kV/m