Which constants and directions always apply when using gravity?

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

3 years ago

You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?

In-s [12.5K]

-- The vertical component of the ball's velocity is 14 sin(51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is 14 cos(51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = 19.56 meters
before it hits the ground.

As usual when we're discussing this stuff, we completely ignore air resistance.

4 0

3 years ago

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During which time does Jamal have the greatest momentum?

Yuliya22 [10]

Answer:

Momentum Packet Answer KEY - Science Online

YOU WILL SEE THE PDF

4 0

2 years ago

SP 15The magnetic field in a region of space is measured to be:This field is known to be caused by a cluster of long-straight wi

tino4ka555 [31]

Answer:

i = 0.477 10⁴ B

the current flows in the counterclockwise

Explanation:

For this exercise let's use the Ampere law

∫ B . ds = μ₀ I

Where the path is closed

Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.

From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.

We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field

Segment on the y axis

L₀ = (y2-y1)

L₀ = 3-0 = 3 cm

Segment on the point x = 2 cm

L₁ = 3-0

L₁ = 3cm

B L = μ₀ I

B 2L = μ₀ I

i = 2 L B /μ₀

i= 2 0.03 / 4π 10⁻⁷ B

i = 4.77 10⁴ B

The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise

8 0

3 years ago

Thermal equilibrium is reached when______________.

Art [367]

two objects in contact with each other are the same temperature

4 0

3 years ago

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Which constants and directions always apply when using gravity?​

Which constants and directions always apply when using gravity?