Solution:
The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :
Differentiating on both the sides with respect to time, we get
Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :
V/m-s
Answer:
The anchor points are 78.37 ft and 111.99 ft
Explanation:
If you look at the attached (Fig 1) you will find that the union of antenna and its guy wires forms two right triangles. To solve problems that involve this kind of triangles, you can apply trigonometric functions (sine, cosine, etc) and Pythagoras Theorem. Trigonometric functions states the relation between angles, sides and hypotenuse of a right triangle. If you look Fig2, considering α angle, "b" is the opposite side, "a" the adjacent side and "c" the hypotenuse. Then
a) Sine (α) = b/c it means opposite side/hypotenuse
b) Cosine (α)= a/c, it means adjacent side/hypotenuse
c) Tangent (α) = b/a opposite side/adjacent side.
Pythagoras theorem states that if you called "a" and "b", the sides of the right triangle, and "c" the hypotenuse, then:
a² + b² = c²
As the problem states the lengths 150 ft and 170 ft represents the value of the hypotenuse of each triangle and 65° is one of the angles of the triangle with 150 ft hypotenuse. So you can solve this using sin (65°) to find the height of the antenna (h) and then the two distances (x and y,).
Sine (65°) = h/ 150 ft ⇒ Sine (65°) x 150ft = h ⇒ h = 127.9 ft.
To find x : Cosine (65°) = x/ 150 ft ⇒ Cosine (65°) x 150 ft = x
⇒ x = 78.37 ft.
And finally, to find y we can apply Pythagoras theorem
(170 ft)² = (127.9 ft)² + y² ⇒ y² = (170 ft)² - (127.9 ft)² ⇒ y = 111.99 ft
Summarizing, the anchor points are 78.37 ft and 111.99 ft
