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3 years ago

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A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p

ositive test charge (q0 = +4.56 x 10-11 C) moves from surface A to another equipotential surface B, which has a radius rB. The work done by the electric force as the test charge moves from surface A to surface B is WAB = -6.95 x 10-9 J. Find rB.

1 answer:

ella [17]3 years ago

7 0

Answer:

$r_B = 1.88 m$

Explanation:

As we know that work done by electric force is given as

$W_e = -q\Delta V$

so here we know that charge is moving from

$r_A = 2.97 m$

to another position

so we will have

$W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}$

$-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})$

$-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})$

$r_B = 1.88 m$

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Answer:

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Explanation:

From the question, we have;

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By plugging in the values, we have;

$\therefore{\Delta f} = 3190 \ Hz = \dfrac{2 \cdot v_{baseball}}{3.0 \times 10^8 \ m/s} \times 2.41 \times 10^{10} \ Hz$

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Sound waves cannot travel in the vacuum of space because there is no medium to transmit these mechanical waves.

On the other hand electromagnetic waves don't require medium for its propagation.

An easy example would be light which is an EM wave reaches earth even though space has no medium.

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