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Bas_tet [7]
3 years ago
14

Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces

. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then 6.3 V, what is the electric field in the region between the plates
Physics
2 answers:
Flauer [41]3 years ago
7 0

Answer:

787.5 V/m

Explanation:

the formula to find the electric field in the region between the plates is

E = V/d

where

V=the voltage supplied by the battery (potential)

d=distance between the plates.

E=electric field

Given:

V= 6.3 V

d= 1.6cm/2=> 0.016m/2

E= V/d=> 6.3/(0.016/2)

E= 787.5 V/m

Therefore, the electric field in the region between the plates is 787.5 V/m

lbvjy [14]3 years ago
6 0

Answer:

The electric field in the region between the plates is 787.5 V/m

Explanation:

The magnitude of the potential difference between the plates is equal to the product of the magnitude of the electric field and the distance between the plates.

The relationship:

E=\frac{V}{d}

Electric field:   E is unknown

Voltage supplied:   V=6.3volts

distance between the plates:  d=\frac{16cm}{2} =\frac{0.016m}{2}=0.008m

E=\frac{V}{d}

Now substitute the values

E=\frac{6.3V}{0.008m} \\E=787.5V/m

The electric field in the region between the plates is 787.5 V/m.

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Answer:

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3 years ago
A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

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m×(v^2)/(R) = μ×m×g

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4 0
3 years ago
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=
zalisa [80]

Answer:

4.24nm

0.385eV

Explanation:

Maximum wavelength (λmax) :

λmax = ( hc) /Φ

h = plancks constant = 6.63 * 10^-34

c = speed of light = 3*10^8

1ev = 1.6 * 10^-19

Φ = 2.93eV = 2.93* (1.6*10^-19) = 4.688*10^-19

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λmax= 4.24nm

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answer:

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