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swat32
3 years ago
6

In deriving the axial electric field for the ring-shaped charge distribution and the electric field from a long line of charge,

the component perpendicular to the resulting field is zero because of what physical property? In deriving the axial electric field for the ring-shaped charge distribution and the electric field from a long line of charge, the component perpendicular to the resulting field is zero because of what physical property? positive and negative charges cancel symmetry only net charge is important superposition integration
Physics
1 answer:
Rudiy273 years ago
5 0

Answer:

d)symmetry

Explanation:

Here is the complete question

In deriving the axial electric field for the ring-shaped charge distribution and the electric field from a long line of charge, the component perpendicular to the resulting field is zero because of what physical property? Answer a)integration b)superposition c)only net charge is important d)symmetry e) positive and negative charges cancel

Solution

The perpendicular components of the resulting electric field is zero because of symmetry since the charge is evenly distributed around the ring or line of charge. At one point, there is an electric field dE due to a small charge element dq and a corresponding electric field dE' due to a small charge element dq' opposite to dq.

Since the charges are symmetrical about the center, the horizontal (or perpendicular) components of the electric fields would cancel out causing the horizontal component of the resulting electric field to be zero.

So, the perpendicular component of the resulting electric field is zero due to symmetry.

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Answer:

Explanation:

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mass = m

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Case 2:

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3 years ago
Question #4
IrinaVladis [17]
2) transverse



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A 2.02 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.22 kΩ resistor. (a) Calculate the current in the
Mandarinka [93]

Answer:

a) 0.048A

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Explanation:

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i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\

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a)

i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A

b)

q=q_o*e^{-\frac{t}{\tau}}

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c) the maximum current occurs when t=0

i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A

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The formula for calculating drag force is as below:

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So, the equation can be,

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Thus, if the speed of the bicycle is doubled, the drag force will get increased by four times.

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