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Sauron [17]
3 years ago
15

Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) what is the electric

field strength between them, if the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 v?
Physics
1 answer:
Alex_Xolod [135]3 years ago
8 0

The expression for the magnitude of the electric field between two uniform conducting plates is

E=\frac{V}{d}

Here, V is potential difference between plates and d is separation between plates.

As the potential 6.00 cm from the zero volt plate (and 4.00 cm from the other) is 420 V.

Therefore,

E = \frac{420 \ V}{6 \times 10^{-2} m } = 70 \times 10^2 \ V/m

Thus, the electric field strength between the plates is 7000 V/ m

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Answer ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀​
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\sf \huge \purple{ Question : -  }

Initial velocity of a car is 36 km/h . Find the distance after min, if it goes with acceleration 2 m/s².

\sf \huge \color {gold}{ AnSwer :- }

Initial velocity, u = 36 km/h

\sf u = 36 \times  \dfrac{5}{18}  {ms}^{ - 1}  = 10 {ms}^{ - 1}

Time, t = 1min

  • t = 60s

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Apply 2nd equation of motion

\longrightarrow \sf s = ut +  \dfrac{1}{2} at ^{2}  \\  \\  \longrightarrow \sf s =(10)(60) +  \frac{1}{2} (2)(60) {}^{2}  \\  \\  \longrightarrow \sf s =600 + 3600   \\  \\  \longrightarrow \sf  \blue{ s =4200m}  \:  \: \large \blue \star

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