Answer:
a) V(r) = k*λ*r^2/R^2 r =< R
V(r) = -2*k*λ*Ln|r/R|
Explanation:
Given:
- The derived results for Electric Fields are:
r =< R, E(r) = 2*k*λ*r / R^2
r > R, E(r) = 2*k*λ/ r
Find:
-Expressions for the electric potential V as a function of r, both inside and outside the cylinder.
Solution:
- From definition we can establish the relation between E(r) and V(r) as follows:
E(r) = - dV / dr
- We will develop expression for each case as follows:
Case 1: r =< R
E(r) = 2*k*λ*r / R^2 = - dV / dr
Separate variables:
2*k*λ*r . dr / R^2 = -dV
Integrating both sides:
2*k*λ/R^2 integral(r).dr = -integral(dv)
k*λ*r^2/R^2 | = - ( 0 - V)
Put limits ( 0-r)
V(r) = k*λ*r^2/R^2
Case 2: r >= R
E(r) = 2*k*λ/ r = - dV / dr
Separate variables:
2*k*λ.dr/ r = -dV
Integrating both sides:
2*k*λ integral(1/r).dr = -integral(dv)
2*k*λ*Ln|r/R| = - ( V - 0)
Put limits ( R -> r)
V(r) = -2*k*λ*Ln|r/R|
C. Democritus was a Greek philosopher who defined the concept of atoms and is the only one of the answer choices who lived during that time period.
Answer:
350 kg m/s
Explanation:
Momentum = mass × velocity
p = mv
p = (100 kg) (3.5 m/s)
p = 350 kg m/s
Fission is used to make the low cost energy
Answer:
depends on the voltage of battery
Explanation:
for example if you connect a battery of 6V,6V will be provided
