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Elena-2011 [213]
3 years ago
15

Calculate the electric field (in V/m) between two parallel plates connected to a 45 V battery if the plates are separated by a)

3 cm and b) 5cm. What is the resulting force on an electron in each of these fields?
Physics
2 answers:
vodka [1.7K]3 years ago
8 0

Answer:

Explanation:

Voltage, V = 45 V

The relation between the electric field and the distance is

E = V / r

(a) r = 3 cm = 0.03 m

E = 45 / 0.03

E = 1500 V/m

(b) r = 5 cm = 0.05 m

E = 45 / 0.05

E = 900 V/m

Sonja [21]3 years ago
6 0

Answer:

Explanation:

Given

Voltage of battery V=45\ V

electric field(E) is given by  

E=|-\frac{dV}{dx}|

i.e. Change in Potential over a distance x

(a)When Plate are separated by 3 cm apart

E=|\frac{45}{3\times 10^{-2}}|

E=1500\ V/m

(b)When Plates are 5 cm apart

E=|-\frac{45}{5\times 10^{-2}}|

E=900\ V/m

Force on electron is given by

F=charge\times Electric\ Field

F=q\times E

For case a

F=1.6\times 10^{-19}\times 1500

F=24\times 10^{-17}\ N

(b)F=1.6\times 10^{-19}\times 900=14.4\times 10^{-17}\ N

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An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
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Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

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a)

How fast will he be moving backward just after releasing the ball?

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m₁v₁ = m₂v₂

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b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

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