Physics College

An isolated charged conductor sphere of radius 12cm has an electric field 49000 N/C at distance 21cm its center
What is its surface charge denisty

Answers

Answer 1

The surface charge density of the isolated charged conductor sphere is approximately 22.15 [tex]C/m^{2}[/tex]

The surface charge density of an isolated charged conductor sphere can be determined using the formula:

σ = E / ([tex]4πr^{2}[/tex])

where σ is the surface charge density, E is the electric field strength, and r is the distance from the center of the sphere.

In this case, the electric field strength E is given as 49000 N/C, and the distance from the center of the sphere r is 21 cm (which can be converted to meters as 0.21 m). Plugging these values into the formula, we have:

σ = (49000 N/C) / [tex](4π * (0.21 m)^{2})[/tex]

Calculating this expression, we find:

σ ≈ 22.15 [tex]C/m^{2}[/tex]

Therefore, the surface charge density is approximately 22.15 [tex]C/m^{2}[/tex]. This indicates the amount of charge per unit area distributed on the surface of the sphere.

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Related Questions

1. Which of the following does not affect the resistance of a wire?
a) Length
b) Temperature
c) Usage time
d) Cross-sectional area
2. If a 12V battery is passing current through a resistor with a current of 2A, what is the value of the resistor?
a 24resistance
b) 14resistance
c) 10resistance
d) 6resistance
3. Describe the differences between series and parallel circuits.
4. A circuit contains resistors of 8resistance and 4resistance，what is combined resistance if the resistors are combined:
a) In series
b) In parallel
5. A 0.5A current is passing across three resistors of 8resistance, 4resistance and 12resistance that are linked in series.
What is the potential difference of the circuit?
6. Wire A has a resistance of 24resistance. If wire B is double the length and has a diameter four times as large as wire A, what is the resistance of wire B?

Answers

The answer right there is b

A particle of mass 2unit moves along space curve defined by ~r(t) = (4t 2 − t 3 )ˆi − 5tˆj + (t 4 − 2)ˆk. Find the force acting on it at any time to work out​

Answers

The force acting on the particle at t = 1 s is 24.4 N.

Mass of the particle, m = 2 units

Distance of the curve, r(t) = (4t²- t³)i - 5tj + (t⁴- 2)k

The velocity of the particle,

v = d[r(t)]/dt = (8t - 3t²)i - 5j + 4t³k

The acceleration of the particle,

a = d²[r(t)]/dt² = (8 - 6t)i + 12t²k

Let the time for which the force is acting be 1 second.

Therefore, acceleration at t = 1 is,

a₁ = 2i + 12k

Hence, the magnitude of acceleration,

|a₁| = √(2²+ 12²)

|a₁| = √148

|a₁| = 12.2 unit/s²

Therefore, the force acting on the particle at t = 1 s is,

F₁ = m x |a₁|

F₁ = 2 x 12.2

F₁ = 24.4 N

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Waves portfolios I will give brainliest!

Now, leave the frequency constant, but play with the amplitude. What is the affect of changing the
amplitude on the tone generated?

Answers

Now, leave the frequency of wave constant, but play with the amplitude. then the effect of changing the amplitude on the tone generated is its loudness. more the amplitude more it will be loud.

Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.

Wavelength is the distance between two points on the wave which are in same phase. Phase is the position of a wave at a point at time t on a waveform. There are two types of the wave longitudinal wave and transverse wave.

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