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An infinite line charge produces a field...

An infinite line charge produces a field of `9xx10^(-4)NC^(-1)` at a distance of 2 cm. Calculate the linear charge density.

A

`10muC//m`

B

`12muC//m`

C

`16muC//m`

D

`18muC//m`

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The correct Answer is:
To solve the problem of finding the linear charge density (λ) of an infinite line charge that produces an electric field (E) of \(9 \times 10^{-4} \, \text{N/C}\) at a distance (r) of 2 cm, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the formula for the electric field due to an infinite line charge**: The electric field (E) produced by an infinite line charge is given by the formula: \[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2\lambda}{r} \] where: - \(E\) is the electric field, - \(\lambda\) is the linear charge density, - \(r\) is the distance from the line charge, - \(\epsilon_0\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)\). 2. **Rearrange the formula to solve for linear charge density (λ)**: We can rearrange the formula to isolate λ: \[ \lambda = \frac{E \cdot r \cdot 4 \pi \epsilon_0}{2} \] 3. **Convert the distance from centimeters to meters**: The distance given is 2 cm. We need to convert this to meters: \[ r = 2 \, \text{cm} = 0.02 \, \text{m} \] 4. **Substitute the known values into the equation**: Now we can substitute the known values into the rearranged formula: - \(E = 9 \times 10^{-4} \, \text{N/C}\) - \(r = 0.02 \, \text{m}\) - \(\epsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)\) Plugging in these values: \[ \lambda = \frac{(9 \times 10^{-4}) \cdot (0.02) \cdot (4 \pi \cdot 8.85 \times 10^{-12})}{2} \] 5. **Calculate the value of λ**: First, calculate \(4 \pi \epsilon_0\): \[ 4 \pi \epsilon_0 \approx 4 \cdot 3.14 \cdot 8.85 \times 10^{-12} \approx 1.11 \times 10^{-10} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \] Now substitute this back into the equation: \[ \lambda = \frac{(9 \times 10^{-4}) \cdot (0.02) \cdot (1.11 \times 10^{-10})}{2} \] \[ \lambda = \frac{(1.998 \times 10^{-15})}{2} \approx 9.99 \times 10^{-16} \, \text{C/m} \] 6. **Final result**: The linear charge density (λ) is approximately: \[ \lambda \approx 1.0 \times 10^{-7} \, \text{C/m} \]

To solve the problem of finding the linear charge density (λ) of an infinite line charge that produces an electric field (E) of \(9 \times 10^{-4} \, \text{N/C}\) at a distance (r) of 2 cm, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the formula for the electric field due to an infinite line charge**: The electric field (E) produced by an infinite line charge is given by the formula: \[ E = \frac{1}{4 \pi \epsilon_0} \cdot \frac{2\lambda}{r} ...
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