Auniformly charged disk with radius r = 40.0 cm and uniform charge density σ = 7.20 ✕ 10−3 c/m2 lies in the xy-plane, with its center at the origin. what is the electric field (in mn/c) due to the charged disk at the following locations? (a) z = 5.00 cm mn/c (b) z = 10.0 cm mn/c (c) z = 50.0 cm mn/c (d) z = 200 cm mn/c

(a) 356.313 MN/C

(b) 307.909 MN/C

(c) 89.078 MN/C

(d) 7.728 MN/C

Explanation:

where;

is Electric field due to location "Z"

K is coloumb's constant = 8.99 X 10⁹ Nm²/C²

is charge density, = 7.20 X 10⁻³ C/m²

R is the radius of the charged disk = 40.0cm

Step 1: calculate the electric field due to location 5.0 cm

= 356.313 MN/C

Step 2: calculate the electric field due to location 10.0 cm

= 307.909 MN/C

Step 3: calculate the electric field due to location 50.0 cm

= 89.078 MN/C

Step 4: calculate the electric field due to location 200.0 cm

= 7.728 MN/C

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

b. a lightning bug, plug of grass, and a worm