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2 years ago

Two point charges, 1.8 pC and −1.8 pC, are separated by 7 µm. What is the dipole moment of this pair of charges?

Physics

1 answer:

Salsk061 [2.6K]2 years ago

3 0

Answer: 12,600,000Cm

Explanation:

From the data's;

Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C

Distance = 7 micrometer, is equal to 0.0000070m

From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.

Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm

NB: The charges are identical.

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Is the energy of a wave is calculated by squaring the frequency of the wave?

Misha Larkins [42]

Nope.
Energy is directly proportional to frequency. and when you calculate energy, you multiply frequency with a constant number called "Planck's Constant"

E = hf

Hope this helps!

5 0

2 years ago

A stone is thrown vertically upwards with an initial velocity of 20m/sec. Find the maximum height ot reaches and the time taken

MAXImum [283]

Answer:

The height reached is 20m, The time taken to reach 20m is 2 seconds

Explanation:

Observing the equations of motion we can see that the following equation will be most helpful for this question.

$v^{2} = u^{2} + 2as$

We are given initial velocity, u

We know that the stone will stop at its maximum height, so final velocity, v

Acceleration, a

And we are looking for the displacement (height reached), s

Substitute the values we are given into the equation

$0^{2} = 20^{2} + 2(10)s$

Rearrange for s

$0^{2} -20^{2} =20s$

$-400=20s$

$\frac{-400}{20} =s$

s = -20 (The negative is just showing direction, it can be ignored for now)

The height reached is 20m

Use a different equation to find the time taken

$s = vt - \frac{1}{2} at^{2}$

Substitute in the values we have

$-20=(0)t - \frac{1}{2} (10)t^{2}$

Rearrange for t

$-20 =0 -5 t^{2}$

$\frac{-20}{-5} =t^{2}$

$4 = t^{2}$

t = 2s

The time taken to reach 20m is 2 seconds

4 0

2 years ago

What type of device forms images through refraction?

Tcecarenko [31]

Answer:

A lens

Explanation:

A lens forms images when light passes Through it bending the rays of in the process.A phenomena called refraction and the speed of light changes in the process because it enters a medium since it's wavelength is reduced.

3 0

1 year ago

What is a charge-coupled device (CCD), and how is it used in astronomy?

Sedaia [141]

Answer:

Charge-coupled device (CCD) is a device that receives and transfers an electrical charge to the next region

Explanation:

Charge-coupled device (CCD) is a device that receives and transfers an electrical charge to the next region where it can be modified like changing it to a electronic value.

In astronomy, high-powered telescopes can be used with CCD device image sensor cameras. The imaging system can concentrate for a number of hours on one place in space once the Earth's rotation synchronizes with the telescope.

6 0

3 years ago

Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25

mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp = 1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0

2 years ago