Explanation:
(a) E = F/q
E = 4.8×10^-17/1.6×10^-19
E = 300 N/C
(b) same magnitude of electric field is exerted on proton
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q =
Therefore, substitute the values into the above formula as follows.
Q =
=
=
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
Answer:
energy which a body possesses by virtue of being in motion.
Explanation:
To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.
They are expressed as,
Where,
= mass of the skier
= mass of the cat
= initial velocity of skier
= initial velocity of cat
= final velocity of both
Re-arrange to find V_f we have,
Once the final velocity is found it is possible to calculate the change in kinetic energy, so
Therefore the amount of kinetic energy converted in to internal energy is 819J