Question

Q1.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

Step-by-step explanation:

Given:

Two point charges 50 μC and 25 μC are 20 cm apart as shown.

To find:

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

Calculation:

Potential energy of system when 25 μC is at point B :

$\rm \: PE_{1} = \dfrac{k(q1)(q2)}{d}$

$\rm \implies\: PE_{1} = \dfrac{9 \times {10}^{9} (50 \times {10}^{ - 6} )(25 \times {10}^{ - 6} )}{ \frac{20}{100} }$

$\rm \implies\: PE_{1} = 9 \times {10}^{9} (50 \times {10}^{ - 6} )(25 \times {10}^{ - 6} ) \times 5$

$\rm \implies\: PE_{1} = 56.25 \: joule$

__________________

(AC)² = 20² + 15²

=> AC = √(625) = 25 cm.

Potential energy when 25 μC is at point C :

$\rm \: PE_{2} = \dfrac{k(q1)(q2)}{r}$

$\rm \implies\: PE_{2} = \dfrac{9 \times {10}^{9} (50 \times {10}^{ - 6} )(25 \times {10}^{ - 6} )}{ \frac{25}{100} }$

$\rm \implies\: PE_{2} = 9 \times {10}^{9} (50 \times {10}^{ - 6} )(25 \times {10}^{ - 6} ) \times 4$

$\rm \implies\: PE_{2} = 45 \:Joule$

So, work done is ∆PE

Work = ∆PE

=> Work = PE2 - PE1

=> Work = 45 - 56.25

=> Work = -11.25 Joule

So, work done is -11.25 Joule