Question

A 6 nC point charge is released from rest at a distance of 15 cm from an infinite line of charge of linear charge density 8 nC/m. What is the charge’s kinetic energy when it’s 60 cm from the line?

Answer 1

KE = 1.196 x 10^(-6) J

Step-by-step 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

Answer 2

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:

`E = (2k\lambda)/(r)`

Here,

k = Coulomb Constant

`\lambda` = Linear Charge Density

r = Distance between them

Work done in displacing charge q will be

`W = \int F dr`

`W = \int q E dr`

Integrating between the two points,

`W = 2kq\lambda \int_(0.15)^(0.60)(1)/(r) dr`

`W = 2(9*10^9)(6*10^(-9))(8*10^(-9)) ln ((0.6)/(0.15))`

`W = 1.20*10^(-6)J`

For energy conservation the work done is equivalent to the applied kinetic energy, therefore the value will be
`1.2*10^(-6)J`