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evablogger [386]
2 years ago
5

A solid nonconducting sphere of radiusRcarries a chargeQdistributed uniformly throughout itsvolume. At a certain distancer1(r1&l

t; R) from the center of the sphere, the electric field has magnitudeE.If the same chargeQwere distributed uniformly throughout a sphere of radius 2R, the magnitude of theelectric field at the same distancer1from the center would be equal to______
Physics
1 answer:
lara [203]2 years ago
3 0

Answer:

E' = \frac{1}{8} E

Explanation:

Given data:

first case

Distance of electric field from center of sphere is r_1 <R

Electric field at r_1< R

E = \frac{kQr_1}{R^3}

second case

Distance of electric field from centre of sphere is r_1 < 2R

Electric field at r_1< 2R

E' = \frac{kQr_1}{8R^3}

so, we have

E' = \frac{1}{8} E

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4 0
2 years ago
Read the following questions and answer them using complete sentences. Be sure to fully explain your answers.
astra-53 [7]

Wave power can be regarded as a reliable source of energy because the ocean currents are always moving.

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6 0
1 year ago
A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 0
3 years ago
A glass capillary tube with a diameter of 8.5 mm and length 8 cm is filled with a salt solution with a resistivity of 2.5 ?m. Wh
MA_775_DIABLO [31]

Answer:

The resistance is 3.5\times10^{-4}\ \Omega

Explanation:

Given that,

Diameter of tube = 8.5 mm

Length = 8 cm

Resistivity = 2.5 m

We need to calculate the resistance

The resistance is equal to the product of the resistivity and length divided by the area of cross section .

In mathematical form,

R = \dfrac{\rho\times l}{A}

Where, \rho=resistivity

l = length

A = area of cross section

Put the value into the formula

R = \dfrac{2.5\times8\times10^{-2}}{3.14\times(\dfrac{8.5}{2}\times10^{-3})^2}

R=3526.32\ \Omega

R=3.5\times10^{-4}\ \Omega

Hence, The resistance is 3.5\times10^{-4}\ \Omega

6 0
3 years ago
José and Laurel measured the length of a stick's shadow during the day. Without knowing the length of the stick, which of their
skelet666 [1.2K]

Answer:

Option B.

Explanation:

Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).

5 0
2 years ago
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