Answer to Question #193045 in Electric Circuits for onia katantumuka

Question #193045

ย A 35 ๐œ‡๐ถ point charge is placed 32 ๐‘๐‘š from an identical 32 ๐œ‡๐ถ charge. How much work would be required to move a 0.50 ๐œ‡๐ถ charge from a point midway between them to a point 12 ๐‘๐‘š closer to either of the charges?


1
Expert's answer
2021-05-13T18:05:47-0400

Solution:-



  • Work is the energy needed to move between the two mentioned points and it is given by

"\\boxed{\n\\begin{aligned}\n\\small W&=\\small q\\Delta V\n\\end{aligned}}"

  • Now q is known & the potential difference between those points are to found to complete the math.
  • Potential of the point midway between the like charges is

"\\qquad\\qquad\n\\begin{aligned}\n\\small V_1&=\\small k\\frac{q_1}{r_1}+k\\frac{q_2}{r_2}\\\\\n&=\\small (9\\times 10^9Nm^2C^-2 )\\Bigg[\\frac{35\\times 10^{-6}C}{0.16m}+\\frac{35\\times10^{-6} C}{0.16m}\\Bigg]\\\\\n&=\\small 3.9375\\times 10^6 V\n\\end{aligned}"

  • Potential of the point 12m from each charge

"\\qquad\\qquad\n\\begin{aligned}\n\\small V_2 &=\\small (9\\times10^9)\\Bigg[\\frac{35\\times 10^-6}{0.12}+\\frac{35\\times 10^{-6}}{0.12}\\Bigg]\\\\\n&=\\small 5.25\\times10^6V\n\\end{aligned}"

  • Then the required work is

"\\qquad\\qquad\n\\begin{aligned}\n\\small W&=\\small 0.50\\times 10^{-6}C\\times 5.25\\times 10^6V\\\\\n&=\\small \\bold{2.625J}\n\\end{aligned}"


  • Theย "\\small 0.5\\mu C"ย charge moves perpendicular to both charges

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