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LUCKY_DIMON [66]
3 years ago
15

Two flat conductors are placed with their inner faces separated by 8 mm. If the surface charge density on one of the inner faces

is 33 pC/m2 and the other inner face −33 pC/m2, what is the magnitude of the electric potential difference between the two conductors? Answer in units of V
Physics
1 answer:
miskamm [114]3 years ago
4 0

Answer:

= 0.0298V

Explanation:

Electric field between two conductor = σ / ε₀

σ = 33pC/m²

= 33 × 10⁻¹²C/m²

ε₀ = 8.85 × 10 ⁻¹²C²/Nm²

E = 33 × 10⁻¹² /  8.85 × 10 ⁻¹²

  = 3.7288N/C

potential difference between two conductors = Ed

where d = 8mm = 8 × 10 ⁻³m

V = 3.7288 ×  8 × 10 ⁻³

   = 29.83 × 10 ⁻³

    = 0.0298V

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A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical
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Answer:

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Given

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Substitute into the formula;

T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} =  \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k =  17.71N/m

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3 years ago
If the distance between the center of two objects is quadrupled. The gravitational
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Answer:

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New force will be :

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So, the new force will change by a factor of 16.

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