Answer:
positive
Explanation:
A positively charged aluminum plate has an excess of protons. When looked at from an electron perspective, a positively charged aluminum plate has a shortage of electrons. In human terms, we could say that each excess proton is rather discontented. It is not satisfied until it has found a negatively charged electron with which to co-habitate. However, since a proton is tightly bound in the nucleus of an atom, it is incapable of leaving an atom in search of that longed-for electron. It can however attract a mobile electron towards itself. So when the positively charged aluminum plate is touched to the neutral metal sphere, countless electrons on the metal sphere migrate towards the aluminum plate. There is a mass migration of electrons until the positive charge on the aluminum plate-metal sphere system becomes redistributed. Having lost electrons to the positively charged aluminum plate, there is a shortage of electrons on the sphere and an overall positive charge. The aluminum plate is still charged positively; only it now has less excess positive charge than it had before the charging process began.
references
https://www.physicsclassroom.com/
Answer:
F₁₋₂ = 1.036*10⁻⁵ [N]
Explanation:
In order to solve this problem, we must use the law of universal gravitation, which can be defined by means of the following expression.
![F_{1-2}=G*\frac{m_{1}*m_{2}}{d^{2} }](https://tex.z-dn.net/?f=F_%7B1-2%7D%3DG%2A%5Cfrac%7Bm_%7B1%7D%2Am_%7B2%7D%7D%7Bd%5E%7B2%7D%20%7D)
where:
F₁₋₂ = Force among the two bodies [N]
G = Universal gravity constant = 6.67 *10⁻¹¹ [N*m²/kg²]
m₁ = mass of the first body = 3.5*10 [kg]
m₂ = mass of the second body = 3*10¹² [kg]
d = separation distance = 26000 [m]
Now replacing:
![F_{1-2}=6.67*10^{-11}*\frac{3.5*10*3*10^{12} }{26000^{2} } \\F_{1-2}=1.036*10^{-5}[N]](https://tex.z-dn.net/?f=F_%7B1-2%7D%3D6.67%2A10%5E%7B-11%7D%2A%5Cfrac%7B3.5%2A10%2A3%2A10%5E%7B12%7D%20%7D%7B26000%5E%7B2%7D%20%7D%20%5C%5CF_%7B1-2%7D%3D1.036%2A10%5E%7B-5%7D%5BN%5D)
Answer:
![V = 228\ V](https://tex.z-dn.net/?f=V%20%3D%20228%5C%20V)
Explanation:
given,
charge of two spherical drop = 0.1 nC
potential at the surface = 300 V
two drops merge to form a single drop
potential at the surface of new drop = ?
![V = \dfrac{kq}{r}](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7Bkq%7D%7Br%7D)
![r = \dfrac{9\times 10^9\times 0.1 \times 10^{-9}}{300}](https://tex.z-dn.net/?f=r%20%3D%20%5Cdfrac%7B9%5Ctimes%2010%5E9%5Ctimes%200.1%20%5Ctimes%2010%5E%7B-9%7D%7D%7B300%7D)
r = 0.003 m
volume = ![2 \times \dfac{4}{3}\pi r^3](https://tex.z-dn.net/?f=2%20%5Ctimes%20%5Cdfac%7B4%7D%7B3%7D%5Cpi%20r%5E3)
= ![2 \times \dfac{4}{3}\pi \times 0.003^3](https://tex.z-dn.net/?f=2%20%5Ctimes%20%5Cdfac%7B4%7D%7B3%7D%5Cpi%20%5Ctimes%200.003%5E3)
= 2.612 × 10⁻⁷ m³
![\dfac{4}{3}\pi R^3 = 2.612\times 10^{-7}](https://tex.z-dn.net/?f=%5Cdfac%7B4%7D%7B3%7D%5Cpi%20R%5E3%20%3D%202.612%5Ctimes%2010%5E%7B-7%7D%20)
R = 0.00396 m
![V = \dfrac{kq}{r}](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7Bkq%7D%7Br%7D)
![V = \dfrac{9\times 10^9 \times 0.1 \times 10^{-9}}{0.00396}](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7B9%5Ctimes%2010%5E9%20%5Ctimes%200.1%20%5Ctimes%2010%5E%7B-9%7D%7D%7B0.00396%7D)
![V = 227.27](https://tex.z-dn.net/?f=V%20%3D%20227.27)
![V = 228\ V](https://tex.z-dn.net/?f=V%20%3D%20228%5C%20V)
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.