Answer:3W
If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much work should it take to move three q point charges from infinity to a distance d apart from each other?
A) 2W
B) 3W
C) 4W
D) 6W
Explanation: calculating work done,W, in moving two positive q point charges from infinity to a valued distance d from each other is
W = k(+q)(+q)/ d
k is couloumb's constant
work done in moving 3 equal positive charges from infinity to a finite distance is given by
W₂=W₄=W₆=k(+q)(+q)/ d
Total work done, W' =k(+q)(+q)/ d + k(+q)(+q)/ d + k(+q)(+q)/ d
= W + W + W = 3W
Carbon is the answer to the problem
Hi There,
This is False.
Hope this helped!
Answer:
Wavelength, 
Explanation:
Given that,
Mass of the particle, 
Acceleration of the particle, 
Time, t = 5 s
It starts from rest, u = 0
The De Broglie wavelength is given by :

v = a × t



Hence, this is the required solution.
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C