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saul85 [17]
4 years ago
7

A solid insulating sphere of radius R = 1.0 m that carries a positive charge Q1 = 1.0 mC uniformly distributed over it is concen

tric with a very thin insulating shell of radius 2R that carries the charge Q2 = -1.0 mC. The charge Q2 is distributed uniformly over the shell.
Use Gauss's law to obtain the magnitude of the electric field:

a. E1 inside the sphere, r < R

b. E2 between the sphere and the shell, R < r < 2R

c. E3 outside the shell, r > 2R

d. Plot the electric field in the radial direction on a graph. Be sure to label the actual magnitude of the electric field at r = 0, r= R, and r = 2R.

Physics
1 answer:
RoseWind [281]4 years ago
3 0

Answer: see attach file

Explanation: In order to solve this problem we have used the Gaussian law in the different regions. Details are included in the attach.

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blagie [28]

Answer:

29.4 uN

Explanation:

The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:

F=k\frac{q_{1} q_{2} }{r^{2}}

where k is a proportionality constant known as the Coulomb's law constant. Its value is 9 \times 10^{9} Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC = 4 \times 10^{-9} C

The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.

Using these values in the formula, we get:

F=9 \times 10^{9} \times \frac{4 \times 10^{-9} \times 4 \times 10^{-9}}{0.7^{2}}\\\\ F=2.94\times10^{-7} N\\\\ F=29.4 \times 10^{-6} N\\\\ F=29.4 \mu N

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8 0
4 years ago
A 5 kg toy is tied to a rope where the tension measures 150 N. What is the weight of the object?
elena-s [515]
Formula from physics to get the answer.
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3 years ago
Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate
vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>

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3 years ago
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Vikki [24]

Answer:

4

Explanation:

a=vf-v

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