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Licemer1 [7]
2 years ago
12

How far apart are two conducting plates that have an electric field strength of 4. 4 kv/m between them, if their potential diffe

rence is 15 kv?
Physics
1 answer:
IRINA_888 [86]2 years ago
6 0

Conducting plates will be 3.75 m apart if an electric field strength of 4. 4 kV/m between them, if their potential difference is 15 kV

The potential difference, also referred to as voltage difference between two given points is the work in joules required to move one coulomb of charge from one point to the other. The SI unit of voltage is the volt. Volt Formula.

In a simple parallel-plate capacitor, a voltage applied between two conductive plates creates a uniform electric field between those plates. The electric field strength in a capacitor is directly proportional to the voltage applied and inversely proportional to the distance between the plates.

E = V/d

E = Electric field strength

d = distance between the plates

V = potential difference

Electric field strength = 4 kV/m

Potential difference  =  15 kV

d = V / E =  15 kV / 4 kV/m

  = 3.75 m

To learn more about capacitor here

brainly.com/question/17176550

#SPJ4

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Anna Litical and Noah Formula are experimenting with the effect of mass and net force upon the acceleration of a lab cart. They determine that a net force of F causes a cart with a mass of M to accelerate at 48 cm/s/s. What is the acceleration value of a cart with a mass of 2M when acted upon by a net force of 2F?

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