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lakkis [162]
3 years ago
11

The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be 3 × 10−5 N · m2 /C when th

e area is parallel to the plate. Find the charge density on the sheet.
Physics
1 answer:
Zanzabum3 years ago
5 0

Answer:

\sigma=2.124\times 10^{-13}C/m^{2}

Explanation:

Given:

Electric Flux = 3\times10^{-5}N.m^2/C

Side of sheet = 5cm

Area of the square sheet, A= 5×5 = 25cm²=25×10⁻⁴ m²

Now

the electric flux (Φ) is given as:

\phi =EA

where, E = Electric field

or

E=\frac{\phi}{A}

substituting the values in the above equation, we get

E=\frac{3\times10^{-5}N.m^2/C}{25\times 10^{-4}}

E=0.012N/C

Now the charge density (σ) on a sheet is given as:

\sigma=2\epsilon_oE

where, \epsilon_o = Permittivity of the free space = 8.85×10⁻¹²

substituting the values in the above equation, we get

\sigma=2\times 8.85\times10^{-12}\times 0.012

\sigma=2.124\times 10^{-13}C/m^{2}

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A 0.02kg dart is loaded into a toy spring gun by compressing the spring 6cm. If the spring has a constant of 20 N/m, find… a. Th
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Answer:

a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m

Explanation:

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Work needed to compress the spring = 0.036J

b) the total energy stored in the spring = the work done to compress the spring = 0.036J

c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.

d) 1/2mv^2 = 0.036

mv^2 = 0.036*2

v^2 = 0.036*2 / 0.02 = 3.6

v = √3.6 = 1.897 approx 1.9m/s

e) kinetic energy of the dart = work done against gravity to get the body to height h

Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward

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0.036 / ( 0.02*9.81) = h

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