Ask Question
25 April, 09:29

When a potential difference of 150 V is applied to the plates of parallel-plate capacitor, the plates carry a surface charge density of 30 nc/cm^2. The spacing between the plates (in μm) is: a) 2.2 b) 1.1 c) 4.4 d) 6.6

+2
Answers (1)
  1. 25 April, 09:31
    0
    s = 4.425*10^-6 μm

    Explanation:

    C = Q/V

    but also:

    C = ∈*A/s

    for A being the area of the plates, the:

    Q = 30 nc / (cm^-2)

    then:

    Q = (30*10^-9) * A/V = ∈*A/s

    ∈ = 8.85 pF/m

    that is: (30*10^-9) / V = (8.85*10^-12) / s

    then, s = (8.85*10^-12) (150) / (30*10^-9)

    = 4.425*10^-6 μm

    Therefore, the spacing between the plates is 4.425*10^-6 μm.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “When a potential difference of 150 V is applied to the plates of parallel-plate capacitor, the plates carry a surface charge density of 30 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers