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3 years ago

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A small charged bead has a mass of 1.9 g . it is held in a uniform electric field e⃗ = (200,000 n/c, up). when the bead is relea

sed, it accelerates upward with an acceleration of 15 m/s2 .

1 answer:

dmitriy555 [2]3 years ago

7 0

As we know that when charge is released in electric field

It will have two forces on it

1. electrostatic force

2. gravitational force

now if the ball will accelerate upwards so we can say

net upward force = mass * acceleration

$qE - mg = ma$

$q*200000 - 1.9 * 10^{-3}* 9.8 = 1.9 * 10^{-3}* 15$

now we can find charge q on it by above equation

$q = 0.2356 * 10^{-6} C$

So above is the charge on the particle

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1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

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V = 17.0 volts

now we will have

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R = 0.91 ohm

now we know that

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$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

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Part b)

Now at higher temperature we have

$V = I R$

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R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

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$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

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2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

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