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:
The diagram represents two charges, q1 and q2, separated by a distance d. Which change would produce the greatest increase in the electrical force between the two charges? *
Explanation:
doubling charge q1, only
First establish the summation of the forces acting int the
ladder
Forces in the x direction Fx = 0 = force of friction (Ff) –
normal force in the wall(n2)
Forces in the y direction Fy =0 = normal force in floor (n1)
– (12*9.81) –( 60*9.81)
So n1 = 706.32 N
Since Ff = un1 = 0.28*706.32 = 197,77 N = n2
Torque balance along the bottom of the ladder = 0 = n2(4 m) –
(12*9.81*2.5 m) – (60*9.81 *x m)
X = 0.844 m
5/ 3 = h/ 0.844
H = 1.4 m can the 60 kg person climb berfore the ladder will
slip
Answer:
14cm
Explanation:
Mass per gram of the piece of wire;
2g of the wire is found in 1m
Since
100cm = 1m;
So;
100cm of the wire contains 2g of the wire
To provide 0.28g
Since;
2g of wire is made up of 100cm
0.28g of wire will be contained in
= 14cm
14cm of the wire will contain 0.28g