A parallel plate capacitor above has square plates 10 cm across on each side. The plates are 50 μm apart, and the electric field strength inside the capacitor is 4.0 x 10⁵ V/m.
A. What is the electric potential difference (voltage) across the capacitor?
B. What is the magnitude of the charge on either plate of the capacitor?
C. What is the capacitance of this capacitor?

a) ΔV = 20 V , b) Q = 35.4 10⁻⁹ C, c)  C = 1.77 10⁻⁹ F

Explanation:

a) The electric potential is

                 ΔV = E x

                 ΔV = 4.0 10⁵ 50 10⁻⁶

                  ΔV = 20 V

c) The capacity of a capacitor is

         C = ε₀  A / d

The area of a square plate is

       A = L²

       A = 0.10²

       A = 0.01 m²

       L = 50mm = 50 10⁻⁶ m

Let's calculate

        C = 8.85 10⁻¹²  0.01 /50 10⁻⁶

        C = 1.77 10⁻⁹ F

b)  the charge is

             Q = C ΔV

             Q = 1.77 10⁻⁹ 20

              Q = 35.4 10⁻⁹ C

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance         (d) = 3 feet

time taken to lift                         (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

                  1 h.p   = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

                      h.p = (force × distance)/ time

                            = (10 × 3)/ 10

                            = 30/10

                            = 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length