A transformer increases and decreases voltage.
The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

The distance from the center of the square to one of the corners is 

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.
2) 

![V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a](https://tex.z-dn.net/?f=V_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%5Csqrt2%7D%20%2B%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B-2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%7D%5C%5CV_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B2%5Ctimes10%5E%7B-6%7D%7D%7B0.05%7D%20%28%5Cfrac%7B1%7D%7B%5Csqrt2%7D-1%29%5C%5CV_b%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%20%284%5Ctimes%2010%5E%7B-5%7D%29%28-0.29%29%5C%5CV_b%20%3D%20%28-%5Cfrac%7B2.9%5Ctimes10%5E%7B-6%7D%7D%7B%5Cpi%5Cepsilon_0%7D%29%5Btex%5D%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E3%29%20The%20work%20done%20on%20q3%20by%20q1%20and%20q2%20is%20equal%20to%20the%20difference%20between%20%20energies.%20This%20is%20the%20work-energy%20theorem.%20So%2C%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20U_b%20-%20U_a)


recoils and must be tethered or he's gone.
Answer:
the intensity of the sun on the other planet is a hundredth of that of the intensity of the sun on earth.
That is,
Intensity of sun on the other planet, Iₒ = (intensity of the sun on earth, Iₑ)/100
Explanation:
Let the intensity of light be represented by I
Let the distance of the star be d
I ∝ (1/d²)
I = k/d²
For the earth,
Iₑ = k/dₑ²
k = Iₑdₑ²
For the other planet, let intensity be Iₒ and distance be dₒ
Iₒ = k/dₒ²
But dₒ = 10dₑ
Iₒ = k/(10dₑ)²
Iₒ = k/100dₑ²
But k = Iₑdₑ²
Iₒ = Iₑdₑ²/100dₑ² = Iₑ/100
Iₒ = Iₑ/100
Meaning the intensity of the sun on the other planet is a hundredth of that of the intensity on earth.