Physics, asked by keejustin94, 24 days ago

Two point charges 50 μC and 25 μC are 20 cm apart as shown. The work done to displace 25 μC charge from B point to C will be
–11.25 J

–16.45 J

–9.29 J

3.25 J

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Answers

Answered by pulakmath007
6

SOLUTION

TO CHOOSE THE CORRECT OPTION

Two point charges 50 μC and 25 μC are 20 cm apart as shown. The work done to displace 25 μC charge from B point to C will be

  • –11.25 J

  • –16.45 J

  • –9.29 J

  • 3.25 J

EVALUATION

Here in the given figure ABC is a right angled triangle with sides AB = 20 cm and BC = 15 cm

Then by the Pythagorean Theorem

 \sf{{(AC) }^{2}  = {(AB) }^{2}  + {(BC) }^{2} }

 \sf{ \implies \: {(AC) }^{2}  = {(20) }^{2}  + {(15) }^{2} }

 \sf{ \implies \: {(AC) }^{2}  =400 + 225 }

 \sf{ \implies \: {(AC) }^{2}  =625 }

 \sf{ \implies \: {AC }  =25 }

∴ AC = 25 cm = 0.25 m

AB = 20 cm = 0.20 m

First we find the potential energy for 25 μC charge at the point B

In that case the 25 μC charge is at the point B and 50 μC charge is at the point A

Now potential energy for 25 μC charge at the point B

 \displaystyle \sf{ =  \frac{k \times q_1q_2   }{AB} }

 \displaystyle \sf{ =  \frac{9 \times  {10}^{9}   \times 50 \times  {10}^{ - 6} \times 25 \times  {10}^{ - 6}    }{0.20} } \:  \:  \: J

 \displaystyle \sf{ =  \frac{9 \times 5 \times 25 \times  {10}^{ - 2}   }{0.20} } \:  \:  \: J

 \displaystyle \sf{ =  \frac{9 \times 5 \times 25 }{20} } \:  \:  \: J

 \displaystyle \sf{ =  \frac{9  \times 25 }{4} } \:  \:  \: J

 \displaystyle \sf{ =  56.25 } \:  \:  \: J

Next we find the potential energy for 25 μC charge at the point C

In that case the 25 μC charge is at the point C and 50 μC charge is at the point A

So the potential energy for 25 μC charge at the point C

 \displaystyle \sf{ =  \frac{k \times q_1q_2   }{AC} }

 \displaystyle \sf{ =  \frac{9 \times  {10}^{9}   \times 50 \times  {10}^{ - 6} \times 25 \times  {10}^{ - 6}    }{0.25} } \:  \:  \: J

 \displaystyle \sf{ =  \frac{9 \times 5 \times 25 \times  {10}^{ - 2}   }{0.25} } \:  \:  \: J

 \displaystyle \sf{ =  \frac{9 \times 5 \times 25 }{25} } \:  \:  \: J

 \displaystyle \sf{ =45} \:  \:  \: J

∴ The work done to displace 25 μC charge from B point to C

= Potential energy for 25 μC charge at the point C - Potential energy for 25 μC charge at the point B

= 45 J - 56.25 J

= - 11.25 J

FINAL ANSWER

Hence the correct option is - 11.25 J

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