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siniylev [52]
3 years ago
9

A thin spherical shell of radius 7.6 cm carries a uniform surface charge density of 6.7 times 10-9 C/m2. The magnitude of the el

ectric field (in N/C) at r = 7.8 cm is approximately:___________.
Physics
1 answer:
zhannawk [14.2K]3 years ago
5 0

Answer:

 E= 1.968 10¹¹  N/C

Explanation:

To calculate the electric field of a sphere we can use Gaussian law

        Ф = ∫ E. dA = q_{int}/ ε₀

In this case we must use a Gaussian surface of spherical shape, since the radii of the sphere and the electric field line have been parallel, consequently the scalar product is reduced to an algebraic product.

         The error of a sphere is

          A = 4π π r²

         E (4π r³) = q_{int} / εo                           (1)

to find the charge inside the sera we use the concept of density

         ρ = quint / V

         qint = rho V

the volume of a sphere is

          V = 4/3 pi r³

we substitute in 1

    E 4π r² = ρ 4/3 π r³

     E = ρ / 3ε₀

now let's change the charge density to the charge

      E = Q 3 / 4π r³    r / 3ε₀

       E = (1 / 4πε₀) Q / r²

       E = k Q /r²

calculate

        E = ρr / 3ε₀

       E = 6.7 10⁻⁹  7.8  10⁻² / 3 8.85 10⁻¹²

         E= 1.968 10¹¹  N/C

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Mama L [17]

Step 1: Look in your book or online for the conical pendulum equation.

Step 2: Look at the drawing and see which angle is involved in the equation.

Answer: It's Angle #2 in your drawing.

8 0
3 years ago
A positively charged sphere with a charge of 8Q is separated from a negatively charged sphere -2Q by a distance r. There is an a
djverab [1.8K]

Answer:

Explanation:

For the first case , the expression for electrostatic force can be given by the following .

F = K x 8Q x 2Q / r² where k is a constant .

F = K 16 Q² / r²

When they touch , some charge is neutralized . Net charge remaining

= 8Q - 2 Q = 6 Q

Charge on each sphere = 6Q/2 = 3 Q .

Force between them

F₁ = k 3Q x 3 Q / r² = k 9 Q² / r²

F₁ / F = 9 / 16

F₁ = 9 F / 16 .

4 0
2 years ago
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
2 years ago
Two men standing on the same side of wall and at the same distance from It, such that they are 4oom apart when one fires a gun t
Mamont248 [21]

Answer:

1. 571.43m/s

2. 142.9m and 342.9m

Explanation:

1.Take the difference in time.

1.2-0.7=0.7 seconds

Take the distance between them and divide with differnce in time.

400÷0.7=571.43 seconds.

2.Take the time of the two men and divide by two.

0.5÷2= 0.25 secs

1.2÷2= 0.6 secs

multiply each with the velocity.

0.25×571.43=142.9m

0.6×571.43=342.9m

8 0
2 years ago
While looking at a cliff, you observe that three visible layers of rocks are tilted about 30 degrees. There are four straight ho
xxTIMURxx [149]

Answer:

the principle of original horizontality and the principle of superposition

Explanation:

The <em>principle of horizontality</em> states that layers of sediment are originally deposited horizontally under the influence of gravity.

The <em>principle of superposition</em> states that the oldest layer layer is at the bottom and each layer above it is younger, with the youngest being at the top.

Unconformities help us find the age of different layers. An unconformity is a surface in which no new solid matter is deposited after a long geologic interval. <em>Angular unconformity </em>is a type of unconformity which different kinds of stratum were tilted or folded before deposition of younger layers of solid matter above the unconformity. Once the layers were folded and tilted, the older layers of the solid matter eroded, then the younger layers were deposited on the older layers. There <em>angular unconformity </em>is the contact between young and old layers of solid matter.

Therefore, these two principles therefore describe how the tilted layers are older than horizontal layers.

3 0
3 years ago
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