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3 years ago

A small glass bead has been charged to 8.0 nc. what is the magnitude of the electric field 2.0 cm from the center of the bead?

1 answer:

7 0

<span>Charge of the glass bead Q = 8.0 x 10^-9 C Distance d = 2.0 cm = 0.02 m Coulombs constant K = 8.99 x 10^9 Nm^2/C^2 Electric Field E = k x Q / d^2 = 8.99 x 10^9 x 8.0 x 10^-9 / (0.02)^2 E = 71.92 / 0.0004 = 17.98 x 10^4 The electric field is 1.8 x 10^5 N/C</span>

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3 years ago

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Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

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The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

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Generally the time taken to attain the second amplitude is mathematically represented as

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=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

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Answer

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