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Sav [38]
3 years ago
8

How many different parts parts of the solar system can you see?

Physics
1 answer:
photoshop1234 [79]3 years ago
6 0

Out of the solar system parts, we can see the sun, a few planets, some meteorites and comets.

Rest all is not visible through naked eyes

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Room temperature

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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
two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
Tomtit [17]

The charges have opposite sign and magnitude 6 \mu C

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

q_1 = q_2 = q

So we can rewrite the equation as

F=\frac{kq^2}{r^2}

And solving for q:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

q_1 = 6 \mu C\\q_2 = -6 \mu C

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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From her bedroom window a girl drops a water-filled balloon to the ground, 4.60 m below. If the balloon is released from rest, h
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Known variables
d=4.6m
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d=1/2gt2
4.6=1/2 -9.8 t2
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