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Help me please, A charge Q is uniformly distributed over a long rod AB of length L, as shown in the figure. The electric potential at the poing O lying at a distance L from the end A is :

A charge Q is uniformly distributed over a long rod AB of length L, as shown in the figure. The electric potential at the poing O lying at a distance L from the end A is :

 

  • Option 1)

    \frac{QIn2}{4\pi \epsilon _{0}L}

  • Option 2)

    \frac{Q}{8\pi \epsilon _{0}L}

  • Option 3)

    \frac{3Q}{4\pi \epsilon _{0}L}

  • Option 4)

    \frac{Q}{4\pi \epsilon _{0}LIn2}

 
Answers (2)
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N neha

As we discussed in

Potential Difference -

V_{B}-V_{A}=\frac{w}{q}

-

 

 Charge on the element

dQ = \frac{Q}{L}dx

Potential at 0

    dV = \frac{1}{4\pi\epsilon_o}\frac{dQ}{x} = \frac{1}{4\pi\epsilon_o}\frac{Q}{Lx}dx

\int dV = \int_{L}^{2L}\frac{1}{4\pi\epsilon_o}\frac{Q}{Lx} dx = \frac{1}{4\pi\epsilon_o}\frac{Q}{L}[lnx]_{L}^{2L}

V = \frac{Q ln2}{4\pi\epsilon_o L}

 

 


Option 1)

\frac{QIn2}{4\pi \epsilon _{0}L}

Option 2)

\frac{Q}{8\pi \epsilon _{0}L}

Option 3)

\frac{3Q}{4\pi \epsilon _{0}L}

Option 4)

\frac{Q}{4\pi \epsilon _{0}LIn2}

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