Actually, alpha-particle is nothing but a "Nucleus of a Helium atom"
It is emitted by some radioactive substances, originally regarded as a ray
Hope this helps!
<span>Ναί is yes
</span>xéro is know
The vertical component of
this acceleration will be:
A= mgsin^2 <span> θ</span>
The downward fforce exerted by the hamster on the block and
on the scle will be
F = -ma
The reading will be like this:
Reading = Mg + mg sin^2 θ
= (0.820) (9.8) +
(0.190) (9.8) sin^2 (0)
<span>Reading = 8.036 N</span>
Answer:
Electric Field at a distance d from one end of the wire is ![E=\dfrac{Q}{4\pi \epsilon_0(L+d)d}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%28L%2Bd%29d%7D)
Electric Field when d is much grater than length of the wire =![\dfrac{Q}{4\pi \epsilon_0\ d^2}](https://tex.z-dn.net/?f=%5Cdfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%5C%20d%5E2%7D)
Explanation:
Given:
- Total charge over the length of the wire = Q
- Length of the wire = L
- Distance from one end of wire at which electric field is needed to find=d
Let dE be the Electric Field due to the small elemental charge on the wire at a distance x from the one end of the wire and let
be the charge density of the wire
![E=\dfrac{dq}{4\pi \epsilon_0x^2}](https://tex.z-dn.net/?f=E%3D%5Cdfrac%7Bdq%7D%7B4%5Cpi%20%5Cepsilon_0x%5E2%7D)
Now integrating it over the entire length varying x from x=d to x=d+L we have and replacing
we have
![E=\int\dfrac{\lambda dx}{4\pi \epsilon_0x^2}\\E=\dfrac{Q}{4\pi \epsilon_0 (L+d)(d)}](https://tex.z-dn.net/?f=E%3D%5Cint%5Cdfrac%7B%5Clambda%20dx%7D%7B4%5Cpi%20%5Cepsilon_0x%5E2%7D%5C%5CE%3D%5Cdfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%20%28L%2Bd%29%28d%29%7D)
When d is much greater than the length of the wire then we have
1+\dfrac{L]{d}≈1
So the Magnitude of the Electric Field at point P = ![\dfrac{Q}{4\pi \epsilon_0\ d^2}](https://tex.z-dn.net/?f=%5Cdfrac%7BQ%7D%7B4%5Cpi%20%5Cepsilon_0%5C%20d%5E2%7D)