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bonufazy [111]
3 years ago
7

Two 4.3546 cm x 4.3546 cm square aluminum electrodes, spaced 0.6408 mm apart are connected to a 73.68 V battery. What is the cap

acitance in pico Farads?
Physics
1 answer:
Helen [10]3 years ago
3 0

Answer:

Capacitance, C = 26.1 picofarad

Explanation:

It is given that,

Side of square, x = 4.3546 cm = 0.043546 m

Distance between electrodes, d = 0.6408 mm = 0.0006408 m

Voltage, V = 73.68 V

Capacitance of parallel plates is given by :

C=\dfrac{\epsilon_oA}{d}

C=\dfrac{8.85\times 10^{-12}\times (0.043546)^2}{0.0006408}

C=2.61\times 10^{-11}\ F

or

C = 26.1 picofarad

So, the capacitance of the capacitor is 26.1 picofarad. Hence, this is the required solution.

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Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential differen
Debora [2.8K]

Answer:

increasing the separation between the plates

Explanation:

The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because

Q(charge) = CV  V = VOLTAGE , c = capacitance  

E = 1/2 eAV^2/ D  ( ENERGY STORED )

where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference

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Answer:

When flying the LNAV Approach, the missed approach point (MAP) would be indicated by reaching:

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Explanation:

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Explanation:

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3 years ago
17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106
sleet_krkn [62]

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied /  d

= 5 x 10³  / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than  breakdown strength for air  3.0×10⁶ V/m

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d = 5 x 10³  / 3.0×10⁶

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