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ra1l [238]
3 years ago
12

A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electr

ic field of magnitude 73.9 N/C that is directed at 20 ∘ from the plane of the sheet. Find the magnitude of the electric fluxthrough the sheet.
Physics
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

6.07 Nm²/C

Explanation:

L = length of the flat sheet = 0.400 m

w = width of the flat sheet = 0.600 m

Area of the rectangular sheet is given as

A = L w

A = (0.400) (0.600)

A = 0.24 m²

E = magnitude of the electric field in the region = 73.9 N/C

θ = Angle of the electric field relative to normal to the plane of sheet = 90 - 20 = 70 deg

Magnitude of magnetic flux through the sheet can be given as

Φ = E A Cosθ

Inserting the values

Φ = (73.9) (0.24) Cos70

Φ = 6.07 Nm²/C

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Calculate the density of each ball. Use the formula
Paraphin [41]

Answer:

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Read 2 more answers
A solid sphere of diameter D = 26 cm has a charge of Q = 4 nano-coulombs uniformly distributed on it. Calculate the magnitude of
kozerog [31]

Answer:

E = 4.83  N/ C

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere:

E= \frac{k*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

Where:

K: coulomb constant (N*m²/C²)

a: sphere radius (m)

Q:  Total sphere charge (C)

r : Distance from the center of the sphere to the region where the electric field is calculated (m)

Equivalences

1nC=10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=4nC=4 *10⁻⁹C

D = 26 cm = 26*10⁻²m = 0.26m

a = D/2 = 0.13m

r= R+a = 2.6 m+ 0.13m = 2.73m

Problem development

Magnitude of the electric field at r = 2.73m from the center of the sphere  

r>a , We apply the Formula (1) :

E= \frac{k*Q}{r^{2} }

E= \frac{9*10^{9}*4*10^{-9} }{2.73^{2} }

E= 4.83 N/ C

3 0
3 years ago
Two small insulating spheres with radius 3.50×10−2 m are separated by a large center-to-center distance of 0.555 m. One sphere i
xz_007 [3.2K]

Answer:

E = 7.83 \times 10^5 N/C

Explanation:

Since we know that two sphere is oppositely charged so net electric field at the mid point of two balls will be sum of the electric field due to each ball at the mid point

So we know that

E = \frac{kq_1}{r^2} + \frac{kq_2}{r^2}

here we know that

q_1 = 4.20 \mu C

q_2 = 2.50 \mu C

r = \frac{0.555}{2}

so we have

E = \frac{(9\times 10^9)(4.20 + 2.50) \times 10^{-6}}{0.2775^2}

E = 7.83 \times 10^5 N/C

4 0
3 years ago
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