An airplane is flying through a thundercloud at a height of 2010 m. (A very dangerous thing to do because of updrafts, turbulence, and the p

Question

An airplane is flying through a thundercloud at a height of 2010 m. (A very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If there is a charge concentration of 46.9 C at height 4090 m within the cloud and −59.7 C at height 570 m, what is the magnitude of the electric field E at the aircraft? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of V/m.

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Thành Đạt 3 days 2021-07-22T01:43:08+00:00 1 Answers 0 views 0

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    2021-07-22T01:45:03+00:00

    Answer:

    The value is E =   3.5619 *10^{5}  V/m

    Explanation:

    From the question we are told that

    The height of the airplane is h  =  2010 \  m

    The first charge concentration is Q  =  46.9  \ C

    The height of the first charge concentration is h_1  = 4090 \  m

    The second charge concentration is Q_2  =  −59.7  \ C

    The height of the second charge concentration is h_1  = 570  \ m

    The electric field due to the first charge concentration is

    E_1 =    \frac{k * |Q_1|}{(h_1 -h)^2}

    Here k is the Coulomb constant given in the question

    E_1 =    \frac{8.98755*10^9 * |46.9|}{(4090 - 2010)^2}

    E_1 =    9.7429 *10^{4} \  V/m

    The electric field due to the second charge concentration is

    E_2 =    \frac{k * |-59.7|}{(h -h_2)^2}

    E_2 =    \frac{8.98755*10^9 * |-59.7|}{(2010 - 570)^2}

    E_2 =    2.5876 *10^{5}  V/m

    Generally the superposition principle can be applied in this question as follows

    E =  E_1 +  E_2

    => E =   9.7429 *10^{4}+  2.5876 *10^{5}

    => E =   3.5619 *10^{5}  V/m

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