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Flura [38]
2 years ago
7

The field between two charged parallel plates is kept constant. If the two plates are brought closer together, the potential dif

ference between the two plates.
Physics
1 answer:
Komok [63]2 years ago
6 0

Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases

The relation between potential difference and the electric field is given by ΔV = E.d

Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.

The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.

The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.

Refer to more about the potential difference here

brainly.com/question/12198573

#SPJ4

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3 years ago
Two 2.0-cm-diameter insulating spheres have a 6.60 cm space between them. One sphere is charged to + 76.0 nC , the other to - 30
e-lub [12.9K]

Answer:

5.2\times 10^5N/C

Explanation:

Since the two charged bodies are symmetric, we can calculate the electric field taking both of them as point charges.

This can be easily seen if we use Gauss's law, \int{E} \, dA=\frac{Q_{enclosed}}{\epsilon_o}

We take a larger sphere of radius, say r, as the Gaussian surface. Then the electric field due to the charged sphere at a distance r from it's center is given by,

E=\frac{1}{4\pi r^2} \frac{Q_{enclosed}}{\epsilon_o}

which is the same as that of a point charge.

In our problem the charges being of opposite signs, the electric field will add up. Therefore,

E_{total}=\frac{1}{4\pi\epsilon_o}\frac{q_1+q_2}{r^2}= (9\times10^9) \frac{(76+30)\times10^{-9}}{((1+3.3)\times10^{-2})^2}N/C =5.2\times10^5N/C

where, r = distance between the center of one sphere to the midpoint (between the 2 spheres)

8 0
3 years ago
Someone, please help!! cant understand
ch4aika [34]

Answer:

4400 N

Explanation: (c)

P= \frac{F}{A}

F= PA

P= pressure

A= area

P=800x10^{3\\ and A= 5.5 x 10^{-3}m^{2}

Force F= 800x10^{3}x0.0055= 4400 N

force exerted by the gas on the piston is 4400 N

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3 years ago
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4 years ago
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