Answer:
The field will remain the same
Explanation:
This is because electric field given as
E1= kq1/r²
And that of second charge
E² = kq2/r²
Is not affected by the size of the second charge q2
ANSWER
Lift frame :
Initial speed of the coin u=0 m/s
Acceleration a=9.8 m/s
2
Initial height of coin from the floor of elevator h=2.45 m
Time taken by coin to hit the floor T=
g
2H
⟹ T=
9.8
2×2.45
=
2
1
s
Answer:
The charge of each charge is ![3.02\times10^{-19} C](https://tex.z-dn.net/?f=3.02%5Ctimes10%5E%7B-19%7D%20C)
Explanation:
When yo have two charged particles they interact exerting an electrostatic force in the other particles, the magnitude of the electrostatic force between two particles is:
(1)
with q1 and q2 the charges, r the distance between them and k the Coulomb's constant (
)
Because the charges we're dealing are identical positive q1=q2, then (1) is:
![F_{e}=k\frac{\mid q^2 \mid}{r^{2}}](https://tex.z-dn.net/?f=F_%7Be%7D%3Dk%5Cfrac%7B%5Cmid%20q%5E2%20%5Cmid%7D%7Br%5E%7B2%7D%7D%20)
Using the values the problem give us:
![6.7\times10^{-9}=8.98755\times10^{9}\frac{\mid q^2 \mid}{(3.5\times10^{-10})^{2}}](https://tex.z-dn.net/?f=%20%206.7%5Ctimes10%5E%7B-9%7D%3D8.98755%5Ctimes10%5E%7B9%7D%5Cfrac%7B%5Cmid%20q%5E2%20%5Cmid%7D%7B%283.5%5Ctimes10%5E%7B-10%7D%29%5E%7B2%7D%7D%20%20)
solving for q:
![q=\sqrt{\frac{(3.5\times10^{-10})^{2}6.7\times10^{-9}}{8.98755\times10^{9}}}](https://tex.z-dn.net/?f=q%3D%5Csqrt%7B%5Cfrac%7B%283.5%5Ctimes10%5E%7B-10%7D%29%5E%7B2%7D6.7%5Ctimes10%5E%7B-9%7D%7D%7B8.98755%5Ctimes10%5E%7B9%7D%7D%7D%20%20%20)
![q= 3.02\times10^{-19} C](https://tex.z-dn.net/?f=%20q%3D%203.02%5Ctimes10%5E%7B-19%7D%20C)
The spontaneous emission of radiations from an unstable nuclei is known as natural radioactivity. on the other hand, The process of emission of radiations from naturally occurring isotopes when they are bombarded with sub-atomic particles or high levels of X-rays or gamma rays called artificial radioactivity.