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)]
Question 1: C Question 2: B, Hope this Helps!
Answer:
The change on the second particle is
.
Explanation:
The period of revolution of the particle in the magnetic field is given by the formula as follows :

It is given that the magnetic field is uniform. The mass of the second particle is the same as that of a proton but thecharge of this particle is different from that of a proton.

If both particles take the same amount of time to go once around their respective circles. So,

So, the change on the second particle is
.
solution:
radius of steel ball(r)=5cm=0.05m
density of ball =8000kgm
terminal velocity(v)=25m/s^2
density of air( d) =1.29 kgm
now
volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3
density of ball= mass of ball/Volume of ball
or, 8000=m/0.00052
or, m=4.16 kg
weight of the ball (W)= mg=4.16×10=41.6 N
viscous force(F)=6 × pi × eta × r × v
=6×3.14×eta×0.05×25
=23.55×eta
To attain the terminal velocity,
Fiscous force=Weight
or, 23.55× eta = 41.6
or, eta = 1.76
whete eta is the coefficient of viscosity.