Consider a beam of electrons in a vacuum, passing through a very narrow slit of width 2.00μm2.00μm. The electrons then head toward an array of detectors a distance 0.9810 mm away. These detectors indicate a diffraction pattern, with a broad maximum of electron intensity (i. e., the number of electrons received in a certain area over a certain period of time) with minima of electron intensity on either side, spaced 0.494 cmcm from the center of the pattern. What is the wavelength λλlambda of one of the electrons in this beam? Recall that the location of the first intensity minima in a single slit diffraction pattern for light is y=Lλ/ay=Lλ/a, where LLL is the distance to the screen (detector) and aaa is the width of the slit.

The purpose of a lab is to do your "expirements" the importance of the topic is so you can now what the story is about and the last one is because no one is me with my question ( what is 3 over 4 x - 9 = 27 maybe you can )

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge