Answer:
Equal to 5000N
Explanation:
The stress on the material is defined by force per unit of cross-sectional area. So it depends on the force and the diameter of the wire, which is the same for both wires. The material that defines the breaking point, is also the same. Therefore, both wires have their breaking point the same at 5000N. The wire length plays no role in here.
You can even see dust flying off of B it’s B
Ok, I think this is right but I am not sure:
Q = ϵ
0AE
A= π π
r^2
=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
(3x10^6 N/C) =3.3x10^-8 C = 33nC N = Q/e = (3.3x10^-8 C)/(1.60x10^-19 C/electron) = 2.1x10^11 electrons
Answer:
Explanation:
electric field at the location of electron
= 9 x 10⁹ x 7.2 / .03²
= 72 x 10¹² N/C
force on electron = electric field x charge on electron
= 72 x 10¹² x 1.6 x 10⁻¹⁹
= 115.2 x 10⁻⁷ N .
C )
work done = charge on electron x potential difference at two points
potential at .03 m
= 9 x 10⁹ x 7.2 / .03
= 2.16 x 10¹² V
potential at .001 m
= 9 x 10⁹ x 7.2 / .001
= 64.8 x 10¹² V
potential difference = (64.8 - 2.16 )x 10¹² V
= 62.64 x 10¹² V .
work done = 62.64 x 10¹² x 1.6 x 10⁻¹⁹
= 100.224 x 10⁻⁷ J .
D )
There will be no change in the magnitude of force on positron except that the direction of force will be reversed . In case of electron , there will be repulsion and in case of positron , there will be attraction .
Work done in case of electron will be positive and work done in case of positron will be negative .
electric field due to charge will be same in both the cases .
