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malfutka [58]
3 years ago
10

A solid conducting sphere of radius 2.00 cm has a charge of 8.84 μC. A conducting spherical shell of inner radius 4.00 cm and ou

ter radius 5.00 cm is concentric with the solid sphere and has a charge of −2.02 μC. Find the electric field at the following radii from the center of this charge configuration.
Physics
1 answer:
AlladinOne [14]3 years ago
6 0

Answer: The electric field is given in three regions well defined; 0<r<2; 2<r<4; 4<r<5 and r>5

Explanation:  In order to solve this problem  we have to use the gaussian law in the mentioned regions.

Region 1; 0<r<2

∫E.ds=Qinside the gaussian surface/ε0

inside of the solid conducting sphere the elevctric field is zero because the charge is located at the surface on this sphere.

Region 2; 2<r<4;

E.4*π*r^2=8,84/ε0

E=8,84/(4*π*ε0*r^2)

Region 3; 4<r<5

E=0 because is inside the conductor.

Finally

Region 4; r>5

E.4*π*r^2=(8,84-2.02)/ε0

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Mariulka [41]

Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

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    G m M / x² = m dv / dt = m dv/dx  dx/dt

    G M / x² = dv/dx   v

    GM dx / x² = v dv

We integrate

    v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m  and the velocity v = vo and the upper limit x = 2.50 10⁸m  with a velocity of v = 8.50 10³ m/s

    ½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

    72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

    72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3  10⁻⁸)

    72.25 10⁶ - v₀² = -1.213  10⁸

    v₀² = 72.25 10⁶ + 1,213 10⁸

    v₀² = 193.6 10⁶

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6 0
3 years ago
Water is an example of ___________, while hydrogen is an example of __________.
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Are you referring to the fact that water is a compound while hydrogen is an element? If I'm wrong just comment and clarify and I can edit it, I don't even know what kind of unit you're in. :)
3 0
3 years ago
Read 2 more answers
A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from
s2008m [1.1K]

E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0
2 years ago
A physics student tests the theory of projectile motion by leaping off a 225 meter tall building. She runs off the building hori
tamaranim1 [39]
In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,

   d = V₀t + 0.5gt²

Substituting the known values,

  225 = (0 m/s)(t) + (0.5)(9.8)(t²)

Simplifying,
 
   t = 6.776 s

To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.


   range = (12.5 m/s)(6.776 s)

   range = 84.7 m

<em>Answer: 84.7 m</em>

4 0
3 years ago
Find the energy u of the capacitor in terms of c and q by using the definition of capacitance and the formula for the energy in
gtnhenbr [62]

The formula for the energy in a capacitor , u in terms of q and c is q²/2c

<h3>What is the energy of a capacitor?</h3>

The energy of a capacitor u = 1/2qv where

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<h3>What is the capacitance of a capacitor?</h3>

Also, the capacitance of a capacitor c = q/v where

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So, v = q/c

<h3>The formula for energy of the capacitor in terms of q and c</h3>

Substituting v into u, we have

u = 1/2qv

= 1/2q(q/c)

= q²/2c

So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c

Learn more about energy in a capacitor here:

brainly.com/question/10705986

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3 0
2 years ago
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