Two charged metallic spheres of radii, R = 10 cms and R2 = 20 cms are touching each other. If the charge on each sphere is +100 nC, what is the electric potential energy between the two charged spheres?

Answers

Answer 1

Answer:

Answer:

$200* 10^(-6)j$

Explanation:

We have given the radius of first sphere is 10 cm and radius of second sphere is 20 cm

So the potential of first sphere will be greater than the potential of the second sphere, so charge will flow from first sphere to second sphere

Let q charge is flow from first sphere to second sphere and then potential become same

So $V=(K(100-q))/(r_1)=(K* 100)/(r_2)$

200-100=2q+q

$q=(100)/(3)=33.33nC$

So $V=(K(100-q))/(r_1)=(9* 10^(9)* (100-33.33)* 10^(-9))/(10* 10^(-2))=6003V$

We know that potential energy U=qV $=33.33* 10^(-9)* 6003=200* 10^(-6)j$

Answer 2

Answer:

Answer:

The electric potential energy between the two charged spheres is $199.9*10^(-6)\ J$

Explanation:

Given that,

Radius of first sphere $R_(1)=10\ cm$

Radius of second sphere $R_(2)=10\ cm$

Charge Q= 100 nC

We know charge flows through higher potential to lower potential.

Using formula of potential

$V=(k(Q-q))/(R_(1))$ ...(I)

$V=(k(Q+q))/(R_(2))$ ...(II)

From equation (I) and (II)

$(k(Q-q))/(R_(1))=(k(Q+q))/(R_(2))$

Put the value into the formula

$((100-q))/(10*10^(2))=((100+q))/(20*10^(-2))$

$(100-q)*20*10^(-2)=(100+q)*10*10^(-2)$

$q=(1000)/(30)$

$q=(100)/(3)$

$q=33.33\ nC$

So, the potential at R₁ and R₂

Using formula of potential

$V=(k(Q-q))/(R_(1))$

Put the value into the formula

$V=(9*10^(9)(100-33.33)*10^(-9))/(10*10^(-2))$

$V=6000.3\ Volt$

We need to calculate the electric potential energy between the two charged spheres

Using formula of the electric potential energy

$U=qV$

$U=33.33*10^(-9)*6.0003*10^(3)$

$U=199.9*10^(-6)\ J$

Hence, The electric potential energy between the two charged spheres is $199.9*10^(-6)\ J$

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A plane monochromatic electromagnetic wave with wavelength ? = 3 cm, propagates through a vacuum. Its magnetic field is described byB? =(Bxi^+Byj^)cos(kz+?t)

where Bx = 3.3 X 10-6 T, By = 3.9 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.

1)

What is f, the frequency of this wave?

GHz

2)

What is I, the intensity of this wave?

W/m2

3)

What is Sz, the z-component of the Poynting vector at (x = 0, y = 0, z = 0) at t = 0?

W/m2

4)

What is Ex, the x-component of the electric field at (x = 0, y = 0, z = 0) at t = 0?

V/m

5)

Compare the sign and magnitude of Sz, the z-component of the Poynting vector at (x=y=z=t=0) of the wave described above to the sign and magnitude of SIIz, the z-component of the Poynting vector at (x=y=z=t=0) of another plane monochromatic electromagnetic wave propagating through vaccum described by:

B? =(BIIxi^?BIIyj^)cos(kz??t)

where BIIx = 3.9 X 10-6 T, BIIy = 3.3 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.

SIIz < 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)

SIIz < 0 and magnitude(SIIz) = magnitude(Sz)

SIIz > 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)

SIIz > 0 and magnitude(SIIz) = magnitude(

Answers

Final answer:

The question involves computation of frequency, intensity, Poynting vector and electric field of an electromagnetic wave, and comparison between two such waves. The solutions result in approximately: 10 GHz for frequency, 3.07 x 10^-12 W/m^2 for intensity, 1.3 X 10^-19 W/m^2 for the z-component of Poynting vector, and 1.43 V/m for the electric field. Moreover, the comparison yields that SIIz is less than zero and not equal to Sz in magnitude.

Explanation:

The subject of your question relates to

electromagnetic waves

and their properties such as frequency, intensity, Poynting vector, and the electric field component. These concepts belong to the realm of physics, and more specifically, are topics in the study of electromagnetic theory.

To solve your questions:

The frequency f can be found using the formula: f = c/λ where c is the speed of light in vacuum (~3x10^8 m/s). For λ = 3 cm or 0.03 m, computation yields f ≈ 10^10 Hz or 10 GHz.
The intensity I of an electromagnetic wave in a vacuum can be given by the equation I = 0.5*c*ε0*E^2, where E is the electric field amplitude. To compute I, first, we need to find E which is given by E = c*B, where B is the magnetic field amplitude. Here, B is the square root of (Bx^2 + By^2) resulting in approximately 4.77*10^-6 T. Thus, E ≈ 1.43 V/m and solving for I gives us I ≈ 3.07 x 10^-12 W/m^2.
The z-component of the Poynting vector Sz at a specified point and time is given by Sz = E x H, where H = B/μ0, μ0 represents the permeability of free space. At t = 0, Sz = Ex*Hy - Ey*Hx = Ex*By, resulting in Sz ≈ 1.3 x 10^-19 W/m^2.
The x-component of the electric field at t = 0 Ex≈1.43 V/m.
Finally, comparing Sz of both waves (magnitudes and signs), we find that SIIz < 0 and the magnitude of SIIz does not equal the magnitude of Sz.

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Final answer:

The frequency of the wave is 10 GHz. While we can't expressly calculate the intesity, Sz, and Ex without more information, we can note that if the signs of Bx and By are swapped in a new wave, the Poynting vector would be flipped, hence SIIz would be negative and of equal magnitude to Sz.

Explanation:

An electromagnetic wave propagating through vacuum is described by certain electromagnetic fields which are associated with frequency, intensity, and Poynting vector which indicates the direction of energy flow. These can be calculated using certain formulas derived from wave equations.

Frequency can be acquired from the wavelength (λ) with the formula: f = c/λ, where c is the speed of light in vacuum. Using given λ = 3 cm, we get f = 10^10 Hz or 10 GHz.

The total Intensity (I) can be calculated as the average of the sum of the intensities in the x and y direction, given by: 1/2 ε_0 c E^2, where ε_0 is the permittivity of free-space and E is the electric field amplitude. However, more information might be needed to calculate this value. Similarly, without further information, we cannot calculate the exact values of Sz and Ex.

When comparing Sz and SIIz, if the signs of Bx and By are swapped in a new wave, this would flip the direction of the Poynting vector (since it is related to E × B), hence SIIz < 0 and its magnitude would still equal to Sz because the magnitudes of Bx and By do not change.

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The time taken by a mass projected verticallyupwards to reach the maximum height (with air
resistance not neglected) is 10 sec. The time of
descent of the mass from the same height will be

Answers

Answer:

10s

Explanation:

The time to get to the maximum would be the same as the time to get down to the maximum unless somehow gravity’s changes during the duration it goes up to and from maximum height.

Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many more turns should be on the secondary compared with the primary?

Answers

I can't give you the actual number of turns, because it's the RATIO
that counts.

However many turns the primary has, the secondary should have
about TEN TIMES that number. Then the transformer will multiply
the primary voltage by 10 ... 120 volts of AC at the primary will
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Note that it HAS TObe AC. If the transformer is supplied with DC,
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Help me with my physics, please

Answers

The right answer would be

-20t+ 80

In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value:g=9.80m/s (s is squared))

Answers

Answer:

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Explanation:

Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of g = 9.8 m/s²

The difference Δ g = 9.96 -9.72 =0.24 m/s²

$The\ percentage\ difference=(0.24)/(9.72)* 100=2.46\ percentage\n$

For first one :

$Error\ in\ the\ percentage =(9.96)/(9.81)* 100 =101.52\ perncetage$

For second :

$Error\ in\ the\ percentage =(9.72)/(9.81)* 100 =99.08\ perncetage$

The mean g(mean )

$g(mean )=(9.96+9.72)/(2)\ m/s^2\ng(mean)=9.84\ m/s^2$

$The\ percentage=(9.84)/(9.8)* 100=100.40\ percentage$

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Final answer:

The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.

Explanation:

In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:

|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%

The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:

For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%

For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%

The percent error of the mean involves doing the above but using the mean of the experimental measurements:

|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%

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The brakes of a car moving at 14m/s are applied, and the car comes to a stop in 4s. (a) What was the cars acceleration? (b) How long would the car take to come to a stop starting from 20m/s with the same acceleration? (c) How long would the car take to slow down from 20m/s to 10m/s with the same acceleration?