A0.500 a current is charging a capacitor that has circular plates 11.0 cm in radius.(a) if the plate separation is 4.00 mm, what is the time rate of increase of electric field between the plates? (b) what is the magnitude of the magnetic field between the plates 5.00 cm from the center?

(a)

We have:

r = 11.0 cm = 0.11 m is the radius of the plates, so their area is

The plate separation is

d = 4.00 mm = 0.004 m

The potential difference across the plates of the capacitor is

(1)

where Q is the charge and C the capacitance; and since

(E= electric field), we can rewrite (1) as

So, the rate variation of the electric field is given by

Since (the current is the rate of change of the charge).

Substituting all the numbers, we find

(b)

The magnetic field between the plates of the capacitor can be found by using Maxwell's equations, and it is given by

where d is the radial distance from the centre.

In this case, we have

d = 5.00 cm = 0.05 m

Therefore, the magnetic field is

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft