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3 years ago

Which of the following best defines insulator?

Physics

1 answer:

aliya0001 [1]3 years ago

6 0

Answer:

A substance with low ability or no ability to conduct energy.

Such as Rubber,Silicone,Plastic

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Which is a true statement about a series circuit?

vaieri [72.5K]

the answer is B your welcome

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3 years ago

Negatively charged particle that orbits the nucleus

Ede4ka [16]

Answer: Electrons.

Explanation:

Electrons are the negatively charged particles that orbit the nucleus of an atom. Protons, on the other hand, are the positively charged particles that orbit the nucleus of an atom.

in the picture, the protons and neutrons are “in” the nucleus and you’ll also see the the electrons are orbiting all around the nucleus.

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3 years ago

1. The electromagnetic waves described in the reading are different in which two ways:

Kipish [7]

Answer:

D is the answer

Explanation:

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3 years ago

What kind of change is it when a liquid turns into a gas by boiling????????????????????

elixir [45]

It is called vaporization. Vaporization is the phase transition from a liquid to a gas by means of evaporation or boiling. Evaporation occurs at temps below the boiling point and occurs on the liquids surface. Boiling is a rapid vaporization that occurs above the boiling temp and below or at the liquids surface.

5 0

3 years ago

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