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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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Answer:

a) 60 V

b) 125 V

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Explanation:

<u>Given</u>

We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.  

<u>Required </u>

We are asked to calculate the potential at the distances

(a) r = 10.0 cm

(b) r = 20.0 cm

(c) r = 40.0 cm  

<u>Solution</u>

(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by

V = q/4\pi∈_o                       (1)

r is the distance where the potential is measured and the term 1/4\pi∈_o equals  9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m  

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(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by  

V = 1*q/4\pi∈_o*R                (2)    

Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m  

V = 1*q/4\pi∈_o*R

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(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)  

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Answer:

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