Physics, 31.08.2019 01:00 musicaljay8320

Three equal charges of magnitude 'e' are located at the vertices of an equilateral triangle of side 1m. where should you place a charge of 2e so that the 2e' does not experience any force?

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Answer from: babyskitt

We would have to locate the charge at the center of the triangle.

Explanation:

I tried my best to make geometric arguments, you can see the attached picture if it helps.

Let's say the triangle has three corners names 1, 2 and 3 as seen in the image. Let's determine the electric field at the center of the triangle. We can see that the x-component of electric field produced by the charge at 1 and the one produce by the charge at 2 will cancel each other out because they have the same but opposite values. The y components on the other hand are equal and are added. The magnitude of all three electric fields are the same (because each corner is at equal distance from the center and all three charges are equal, see the formula for the electric field: $|E|=\frac{1}{4.\pi.\epsilon_0}\frac{q}{d^2}$ where is the distance between the point of observation and the charge). Thus lets call E the magnitude of the electric field produced by any of the charges.

We can clearly see that the y-component of the electric filed produced by 1 is $E\sin(30)$ , the same goes for the one produced by the charge at 2, knowing that $\sin(30)=\frac{1}{2}$ , we have that the total field due to the charges at 1 and 2 are equal to E in the positive y-direction. The field produced by the third charge is clearly -E in the negative y-direction thus canceling the contribution from 1 and 2. The total electric field at the center of the triangle is zero, this also mean that any charge placed at that point will not feel any force.

Three equal charges of magnitude 'e' are located at the vertices of an equilateral triangle of side

Answer from: Quest

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb

Answer from: Quest

it is given that trillions of microbes make up to 1% to 3% of the body mass.

let's take average of this percentage: 2%

consider the average mass of the human body as 70 kg.

then, trillions of microbes would have mass $= 2\%\hspace{1mm}of \hspace{1mm} 70\hspace{1mm} kg=\frac {2}{100}\times 70 =1.4\hspace{1mm} kg$

therefore, 100 trillion micro-organisms would weigh:

$\rightarrow 100\times1.4 kg = 140 kg$

this is an approximate value and could change depending on the mass of the human body and percentage of the microbes present in it.

average mass per microorganism = $\frac{1.4 kg}{10^12}=1.4\times10^{-12} kg$

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