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3 years ago

A large drill is used to make a hole through rocks in the earths surface to excavate for oil

Physics

1 answer:

Mamont248 [21]3 years ago

3 0

Answer:

$\Delta GPE = mg\Delta h$

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its location in the gravitational field.

This energy is always measured as a difference with respect a "zero level" - generally taken as the ground level.

Therefore, it generally makes more sense to measure the change in gravitational potential energy of an object when its height is changed; this is given by:

$\Delta GPE = mg\Delta h$

where:

m is the mass of the object

$g=9.8 m/s^2$ is the acceleration due to gravity

$\Delta h$ is the change in height of the object

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Which atmospheric condition is most likely responsible for the wind blowing the clouds from the sea towards the land?

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2 years ago

Which one of newton's laws does a doll riding a dog represent

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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.

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3 years ago

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Answer:

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