Q1.Two point charges 50 μC and 25 μC are 20 cm apart as shown. The w

1.	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» 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 : $\rm \: PE_{1} = \dfrac{k(q1)(q2)}{d}$ $\rm \implies\: PE_{1} = \dfrac{9 \times {10}^{9} (50 \times {10}^{ - 6} )(25 \times {10}^{ - 6} )}{ \frac{20}{100} }$ $\rm \implies\: PE_{1} = 9 \times {10}^{9} (50 \times {10}^{ - 6} )(25 \times {10}^{ - 6} ) \times 5$ $\rm \implies\: PE_{1} = 56.25 \: <klux>JOULE</klux>$ __________________ (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

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»

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 :

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

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

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

$\rm \implies\: PE_{1} = 56.25 \: <klux>JOULE</klux>$

__________________

(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