To develop this problem we will apply the concepts related to the electric field and the Force, the equivalence between the charge and the electric field. Defining the electric field then as the product between the Coulomb constant, the linear charge density and the distance between both points, we have:
[tex]E = \frac{2k\lambda}{r}[/tex]
Here,
k = Coulomb Constant
[tex]\lambda[/tex] = Linear Charge Density
r = Distance between them
Work done in displacing charge q will be
[tex]W = \int F dr[/tex]
[tex]W = \int q E dr[/tex]
Integrating between the two points,
[tex]W = 2kq\lambda \int_{0.15}^{0.60}\frac{1}{r} dr[/tex]
[tex]W = 2(9*10^9)(6*10^{-9})(8*10^{-9}) ln (\frac{0.6}{0.15})[/tex]
[tex]W = 1.20*10^{-6}J[/tex]
For energy conservation the work done is equivalent to the applied kinetic energy, therefore the value will be [tex]1.2*10^{-6}J[/tex]
Answer:
KE = 1.196 x 10^(-6) J
Explanation:
The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law;
E = λ/(2πrεo)
Where,
λ is linear charge density = 8 nC/m = 8 x 10^(-9) C/m
εo is electric constant
permittivity of free space
vacuum permittivity with a value of 8.854 × 10^(−12) C²/N.m²
r is distance between both charges.
Now, Work done is given as;
W = Force x Distance = F•r
Now, F = qE
Where q is charge = 6nc = 6 x 10^(-9) C
Now, since we are looking for K.E after 60cm from the line of 15cm,we can express W as;
W = ∫F•dr
Applying to this question, we have;
W = ∫qE•dr at boundary of 0.6m and 0.15m
W = ∫qλ/(2πrεo)•dr at boundary of 0.6m and 0.15m
W = (qλ/2πεo)∫(1/r)•dr at boundary of 0.6m and 0.15m
Integrating and plugging in the relevant values, we have;
W = [6 x 10^(-9) x 8 x 10^(-9)]/(2π x 8.854 × 10^(−12)] [In(0.6) - In(0.15)]
W = 0.8628 x 10^(-6) (1.3863) = 1.196 x 10^(-6) J
From conservation of energy, Work done is equal to kinetic energy.
KE = 1.196 x 10^(-6) J
