To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

Here,
k = Coulomb's constant
q = Charge of proton and electron
r = Distance
Replacing we have that,


The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.
The acceleration of the electron is given as



The acceleration of the proton is given as,



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)


Answer:
Torque is a measure of the force that can cause an object to rotate about an axis.
The relationship is directly proportional; as temperature increases, volume increases in the same way.
Charles's law states that at a constant pressure, the volume of fixed a mass of a gas is directly proportional to its absolute temperature or kelvin temperature.
Mathematically, this law can be written as follows;

This law explains the direct relationship between Volume of the gas and its Kelvin temperature. That is, as Temperature increases, the volume of the gas increases.
Thus, the correct statement is "The relationship is directly proportional; as temperature increases, volume increases in the same way".
Learn more here: brainly.com/question/16927784