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)


Water is usually used to cool down automobile engines when they get hot, yes. Therefore, that means water has a high heat capacity.
That makes the answer letter D you provided above.
D) Water has a high heat capacity.
Another example would be trying to put out a fire with a bucket of water. Usually, you can put out the fire debating on the size!
Answer:


Explanation:
<u>Given Data:</u>
Weight = W = 65 N
Height = h = 2 m
Time = t = 4 secs
<u>Required:</u>
Power = P = ?
Work Done in the form of Potential Energy = P.E. = ?
<u>Formula:</u>
P.E. = Wh
P = P.E. / t
<u>Solution:</u>
P.E. = (65)(2)
P.E = 130 Joules
P = P.E. / t
P = 130 / 4
P = 32.5 Watts
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807 </h3>
2/5 = .4
.4*100= 40%
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