Question

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​

Answer 1

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

Now potential energy for 25 μC charge at the point B

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

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

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

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

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

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

Again 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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