Answer:
E = k Q / [d(d+L)]
Explanation:
As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field
E = k ∫ dq/ r² r^
"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.
Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant
λ = Q / L
If we derive from the length we have
λ = dq/dx ⇒ dq = L dx
We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge
dE = k dq / x²2
dE = k λ dx / x²
Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider
E = k 
We take out the constant magnitudes and perform the integral
E = k λ (-1/x)
Evaluating
E = k λ [ 1/d - 1/ (d+L)]
Using λ = Q/L
E = k Q/L [ 1/d - 1/ (d+L)]
let's use a bit of arithmetic to simplify the expression
[ 1/d - 1/ (d+L)] = L /[d(d+L)]
The final result is
E = k Q / [d(d+L)]
Answer:
Ax = 0
Ay = 6 m
Bx = 8 cos phi = cos 34 = 6.63 m
By = 8 sin phi = 8 sin (-34) = -4.47 m
Rx = Ax + Bx = 0 + 6.63 = 6.63 m
Ry = Ay + By = 6 - 4.47 = 1.53 m
R = (6.63^2 + 1.53^2)^1/2 = 6.80 m
tan theta = Ry / Rx = 1.53 / 6.8 = ,225
theta = 12.7 deg
Answer:let initial velocity u=14m/s
Final velocity v=20m/s
Time taken t=30
Acceleration =a
V=u +at
a= (20-14)/30
a=0.2m/s^2
Explanation:
Acceleration is the change in velocity with respect to time.
Answer:
The distance is 
Explanation:
From the question we are told that
The initial speed of the electron is 
The mass of electron is 
Let 
be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value
Let 
be the initial kinetic energy of the electron \
Let 
be the kinetic energy of the electron at the distance from the proton
Considering that energy is conserved,
The energy at the initial position of the electron = The energy at the final position of the electron
i.e
are the potential energy at the initial position of the electron and at distance d of the electron to the proton
Here 
So the equation becomes
Here 
are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton
is electrostatic constant with value 
i.e 
is the velocity at distance d from the proton = 2
So the equation becomes
![\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv_i%5E2%20%20%3D%204%20%5B%5Cfrac%7B1%7D%7B2%7Dmv_i%5E2%20%5D-%20%5Cfrac%7Bk%28q%29%5E2%7D%7Bd%7D)
![3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}](https://tex.z-dn.net/?f=3%5B%5Cfrac%7B1%7D%7B2%7Dmv_i%5E2%20%5D%20%3D%20%5Cfrac%7Bk%28q%29%5E2%7D%7Bd%7D)
Making d the subject of the formula
