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Gnoma [55]
3 years ago
7

Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 X10^-9 C and the other

is given a charge of -18 X 10^-9 C. a. Find the electric force exerted on one sphere by the other. b. The sphere are connected by a conducting wire. After equilibrium has occurred, find the electric force between the two spheres.
Physics
1 answer:
svlad2 [7]3 years ago
6 0

Answer:

Explanation:

Force between two charged conducting sphere

= k x Q₁ x Q₂ / r² ,  k is a constant  Q₁ and Q₂ are charges and   r is distance between them .

= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²

= 21600 x 10⁻⁹

= 2.16 x 10⁻⁵ N .

b )

After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .

Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2

= - 3 x 10⁻⁹ C .

Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²

= 900 x 10⁻⁹ N .

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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h3>DATA:</h3>

Height of Aeroplane = 500 meters

Speed of Aeroplane = 200 m/s

Acceleration due to gravity = g = 10m/s^2

time of package to reach the ground = t = ?

Horizontal distance = d = ?

<h2>_____________________________________</h2><h3>SOLUTION:</h3>

When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the plane as the plane is just moving in horizontal direction. So the bomb fall and will travel forward.

Initial horizontal velocity is given by,

                                            V_{ix} = 200 ms{-1}

Initial vertical velocity is given by,

                                            V_{iy} = 0m/s

By the second equation of motion under gravity,

                                           H = V_{iy}t + \frac{1}{2}gt^2\\\\500 = (0)t + \frac{1}{2}x10xt^2 \\\\500 = 0 + 5t^2\\\\t^2 = \frac{500}{5}\\\\t^2 = 100\\\\t = 10 seconds

<h2>_____________________________________</h2>

<u>For horizontal distance(d)</u>:

Horizontal distance has no acceleration thus d is given by

                                          d = vt

                                          d = 200x10

                                          d = 2000 meters

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

                 

4 0
3 years ago
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8.04 h + 1.29 = 1.536

h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm

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A uniform electric field exists in the region between two oppositely charged plane parallel plates. a proton is released from re
dedylja [7]

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E is the magnitude of the electric field; e = 1.6 * 10^-19 is the elementary charge; mp = 1.67*10^-27 kg is the proton mass; a is the acceleration.

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The proton accelerated

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From equations (1) and (3)

E= 32.51 V/m

Electric field

The physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field).  It can also refer to a system of charged particles' physical field. Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.

To learn more about an electric field refer here:

brainly.com/question/15800304

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2 years ago
According to the Standard Model of Particle Physics, a neutrino is a type of
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5 0
3 years ago
Read 2 more answers
2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo
seropon [69]

Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=2\pi f

Put the value into the formula

\omega=2\times\pi\times60

\omega=376.9\ rad/s

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

B_{max}=2.901\times10^{-13}\ T

Hence, The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

7 0
3 years ago
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