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:
1. Largest force: C; smallest force: B; 2. ratio = 9:1
Explanation:
The formula for the force exerted between two charges is
where K is the Coulomb constant.
q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.
For simplicity, let's combine Kq₁q₂ into a single constant, k.
Then, we can write
1. Net force on each particle
Let's
- Call the distance between adjacent charges d.
- Remember that like charges repel and unlike charges attract.
Define forces exerted to the right as positive and those to the left as negative.
(a) Force on A
(b) Force on B
(C) Force on C
(d) Force on D
(e) Relative net forces
In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.
2. Ratio of largest force to smallest
Answer:
160N
Explanation: When 80kg mass is one group . It's reaction force acting on a ground.
Weight of the object = 80*10
= 800 N
Here we are given cofficient of static friction its 0.2. It should be smaller than 1
Friction force = Reaction * Friction Cofficient
Reaction = 800N ( Considering Vertical Equilibrium )
F = 800* 0.2
F = 160N
Answer:
physical measurement comprises the measurement of objects, things, etc. and is concerned with the measurement of height, weight, length, size, volume etc.