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?

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

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