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sweet [91]
3 years ago
6

A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

r face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? (ε0 = 8.85 × 10-12 C2/N · m2) A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The upper face carries +95.0 nC/m2 while the lower face carries -25.0 nC/ m2. What is the magnitude of the electric field at a point within the sheet 2.00 cm below the upper face? ( = 8.85 × 10-12 C2/N · m2) 0.00 N/C 6.78 × 103 N/C 7.91 × 103 N/C 1.36 × 104 N/C 3.95 × 103 N/C
Physics
1 answer:
Nonamiya [84]3 years ago
4 0

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

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sleet_krkn [62]

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Resistivity \rho =\frac{RA}{l}

It depends upon cross sectional area and length of material

Explanation:

The resistance of any material is given by R=\frac{\rho l}{A}, here \rho is the resistivity of material , l is length of material and A is cross sectional area

So resistivity \rho =\frac{RA}{l}

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3 0
3 years ago
A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
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Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

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So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

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I hope it helps you!

       

8 0
3 years ago
A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea
fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

Xc = \frac{1}{2\pi fC}

where,

Xc\\ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

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To find,

Xc =?

Xc' = \frac{1}{2\pi f'C}

      = \frac{1}{2\pi 2f C}

      = \frac{1}{2} (\frac{1}{2\pi fC} )\\

Xc' = \frac{1}{2} Xc

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

  1. The capacitive reactance is doubled.
  2. The capacitive reactance is traduced by a factor of 4.
  3. The capacitive reactance remains constant.
  4. The capacitive reactance is quadrupled.
  5. The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

#SPJ4

3 0
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Name any two liquid which are found in our human eyes​
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Aqueous humor and vitreous humor are the liquids present in the human eye.

<em>Hope </em><em>it</em><em> helped</em><em> you</em><em>.</em><em>.</em><em>.</em><em> </em><em>pls </em><em>mark</em><em> brainliest</em>

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