Physics, asked by Blockhead5017, 5 hours ago

When a particle of charge 10microcoulomb brought from infinity to point p 2 my of work is done by the external force what is the potential at p

Answers

Answered by Anonymous
27

Answer:

Potential at a point in a field is defined as the amount of work done in bringing a unit positive test charge from infinity to that point along any arbitrary path. mathematically;

\footnotesize\longrightarrow\:\sf V = \dfrac{W}{q_{0}}

  • Work done = 2 mJ
  • Test charge = 10 μC

\footnotesize\longrightarrow\:\sf V = \dfrac{2 \times  {10}^{ - 3} }{10 \times  {10}^{ - 6} }

\footnotesize\longrightarrow\:\sf V = \dfrac{2 \times  {10}^{ - 3}   \times  {10}^{6} }{10  }

\footnotesize\longrightarrow\:\sf V = \dfrac{2 \times  {10}^{3}    }{10  }

\footnotesize\longrightarrow\:\sf V = 0.2\times  {10}^{3}

\footnotesize\longrightarrow\:\sf V = 2\times {10}^{ - 1} \times    {10}^{3}

\footnotesize\longrightarrow\:\sf V = 2 \times {10}^{2}

\footnotesize\longrightarrow\: \underline{ \underline{\sf V = 2 00 \: J/C}}

MORE INFORMATION:

\footnotesize\maltese \: \:\sf 1 \: mC =  {10}^{ - 3} C

\footnotesize\maltese \: \:\sf 1 \:  \mu C =  {10}^{ - 6} C

\footnotesize\maltese \: \:\sf 1 \: n C =  {10}^{ - 9} C

\footnotesize\maltese \: \:\sf 1 \: mJ  =  {10}^{ - 3} J

• Electric Potential is a scalar quantity.

• S.I unit is Joule/Coulomb (J/C) or Volt (v)

• Dimensional Formula = [V] = [ML²T-³A-¹]

Answered by niltkumarsaha629
2

Answer:

In the question the unit of the work done,is missing.

I assume that it is 2 mJ .

Potential at a specific point V = W/Q [ where V is in volt when W is in Jule and Q is in coulomb]

Here, W = 2mJ = 2 × 10-³ J

Q = 10 microcoulmb= 10 × 10-⁶ C

Therefore, the potential difference at p = 2× 10-³/10×10-⁶ V = 200 V

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