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Electrostatics question

2013 · Shift 0 · Q64
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Electrostatics question

2013 · Shift 0 · Q64

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A charge QQQ is uniformly distributed over a long rod ABABAB of length LLL as shown in the figure. The electric potential at the point OOO lying at distance LLL from the end AAA is JEE Main 2013 (Offline) Physics - Electrostatics Question 222 English
  1. A
    Q8πε0L{Q \over {8\pi {\varepsilon _0}L}}8πε0​LQ​
  2. B
    3Q4πε0L{{3Q} \over {4\pi {\varepsilon _0}L}}4πε0​L3Q​
  3. C
    Q4πε0L ln⁡ 2{Q \over {4\pi {\varepsilon _0}L\,\ln \,2}}4πε0​Lln2Q​
  4. D
    Qln⁡ 24πε0L s{{Q\ln \,2} \over {4\pi {\varepsilon _0}L\,{}^s}}4πε0​LsQln2​
View written solutionFree

Correct answer: D

  1. Set up the charge distribution

A total charge QQQ is uniformly distributed on a rod of length LLL.

So the linear charge density is

λ=QL.\lambda = \frac{Q}{L}.λ=LQ​.
  1. Geometry of the problem

Point OOO lies on the extension of the rod, at a distance LLL from end AAA.

So if we measure distance from OOO toward the rod:

  • the nearer end AAA is at distance LLL from OOO,
  • the farther end BBB is at distance 2L2L2L from OOO.

Thus the rod spans radial distances from r=Lr=Lr=L to r=2Lr=2Lr=2L from point OOO.

  1. Potential due to a small element

Take a small charge element dqdqdq at distance rrr from OOO. Then

dV=14πε0dqr.dV = \frac{1}{4\pi\varepsilon_0}\frac{dq}{r}.dV=4πε0​1​rdq​.

Since dq=λ drdq = \lambda \, drdq=λdr, we get

dV=14πε0λ drr.dV = \frac{1}{4\pi\varepsilon_0}\frac{\lambda \, dr}{r}.dV=4πε0​1​rλdr​.
  1. Integrate over the full rod
V=14πε0∫L2Lλ drr=λ4πε0∫L2Ldrr.V = \frac{1}{4\pi\varepsilon_0}\int_L^{2L} \frac{\lambda \, dr}{r} = \frac{\lambda}{4\pi\varepsilon_0} \int_L^{2L} \frac{dr}{r}.V=4πε0​1​∫L2L​rλdr​=4πε0​λ​∫L2L​rdr​.

Using

∫drr=ln⁡r,\int \frac{dr}{r} = \ln r,∫rdr​=lnr,

we get

V=λ4πε0[ln⁡r]L2L=λ4πε0ln⁡(2LL).V = \frac{\lambda}{4\pi\varepsilon_0} \left[\ln r\right]_L^{2L} = \frac{\lambda}{4\pi\varepsilon_0} \ln\left(\frac{2L}{L}\right).V=4πε0​λ​[lnr]L2L​=4πε0​λ​ln(L2L​).

So,

V=λln⁡24πε0.V = \frac{\lambda \ln 2}{4\pi\varepsilon_0}.V=4πε0​λln2​.

Now substitute λ=Q/L\lambda = Q/Lλ=Q/L:

V=Qln⁡24πε0L.V = \frac{Q\ln 2}{4\pi\varepsilon_0 L}.V=4πε0​LQln2​.
  1. Match with the options

This matches Option D:

Qln⁡24πε0L.\boxed{\frac{Q\ln 2}{4\pi\varepsilon_0 L}}.4πε0​LQln2​​.

(The extra symbol shown in the option denominator appears to be a printing/OCR error; the intended expression is clearly this one.)

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

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