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.

Answer :

Answer:

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

Explanation:

From the question we are told that

The height of the airplane is [tex]h = 2010 \ m[/tex]

The first charge concentration is [tex]Q = 46.9 \ C[/tex]

The height of the first charge concentration is [tex]h_1 = 4090 \ m[/tex]

The second charge concentration is [tex]Q_2 = −59.7 \ C[/tex]

The height of the second charge concentration is [tex]h_1 = 570 \ m[/tex]

The electric field due to the first charge concentration is

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

Here k is the Coulomb constant given in the question

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

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

The electric field due to the second charge concentration is

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

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

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

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

[tex]E = E_1 + E_2[/tex]

=> [tex]E = 9.7429 *10^{4}+ 2.5876 *10^{5} [/tex]

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

Other Questions