Answer:
Explanation:
Electric field due to a charge Q at a point d distance away is given by the expression
E = k Q / d , k is a constant equal to 9 x 10⁹
Field due to charge = 3 X 10⁻⁹ C
E = E = 
Field due to charge = 4 X 10⁻⁹ C
![E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}](https://tex.z-dn.net/?f=E%20%3D%20%5Btex%5D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes4%5Ctimes10%5E%7B-9%7D%7D%7B%282-d%29%5E2%7D)
These two fields will be equal and opposite to make net field zero
=
[/tex]


d = 0.928
A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metalloid. Hope this helps!
The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha
In both cases less energy is required
But comparetively Mg require more energy than K
Let's see the electron configuration of Both
- [Mg]=1s²2s²2p⁶3s²=[Ne]3s²
- [K]=1s²2s²2p⁶3s²3p⁶4s¹=[Ar]4s¹
K has only one valence electron so very less ionization enthalpy so less energy required
Mg has 2 so more IE hence more energy required