Question

Four point charges of charge 1 uc each are placed at four corners of a square of side 2 m. The work done in removing a charge - 2 pc from centre of square to infinity is​

Answer 1

Answer:

36√2 x 10^(-9) J ≈ 5.09 x 10^(-8) J

Explanation:

Workdone in moving a charge(q) from infinity to a certain point(at P.D. V) is qV.

Therefore,

Workdone in moving a charge(q) from a certain point(at P.D. V) to infinity is - qV.

In this condition, r = ½ diagonal = √2 m, and work by each charge is same, so 4 work times of work by any one.

Work = - 4q(k Q/r) = - 4k(qQ/r)

Work = - 4(9 × 10^9) [(-2p)(1u)/√2]

Work = - (36 × 10^9) [- √2 pu]

Work = 36√2 × 10^9 × 10^(-18)

Work = 36√2 × 10^(-9) ≈ 5.09 × 10^(-8) J

*u refers to micro = 10^(-6).

*p refers to pico = 10^(-12), that's how pu = 10^(-18).

*proof for this specific condition:

https://brainly.in/question/17565289

Just substitute your values in

W = 4√2 (kqQ/r)

= 4√2 (9 x 10^9) x (1u)(2p)/2

= 36√2 x 10^9 x 10^(-18)

= 36√2 x 10^(-9) J

Answer 2

Given:

Four point charges of charge 1 uc each are placed at four corners of a square of side 2 m.

To find:

The work done in removing a charge - 2 pc from centre of square to infinity ?

Calculation:

• The work done to move that -2 pC charge to infinity can be found out from the change in potential energy of the system.

At the centre of sphere, net potential energy is:

$\rm \: PE1 =4 \times \dfrac{k(q1)(q2)}{d }$

• Value of d is half of diagonal (i.e. √2 metres).

$\rm \implies\: PE1 =4 \times \dfrac{(9 \times {10}^{9} )( - 2 \times {10}^{ - 12} )(1 \times {10}^{ - 6}) }{ \sqrt{2} }$

$\rm \implies\: PE1 =2 \sqrt{2} (9 \times {10}^{9} )( - 2 \times {10}^{ - 12} )(1 \times {10}^{ - 6})$

$\rm \implies\: PE1 = - 50.9 \times {10}^{ - 9} \: joule$

Now, potential energy after the charge reaches infinity:

$\rm \: PE2=4 \times \dfrac{k(q1)(q2)}{d2 }$

$\rm \implies\: PE2=4 \times \dfrac{k(q1)(q2)}{ \infty }$

$\rm \implies\: PE2=0 \: joule$

So , net work done is :

$\rm \: W = PE2 - PE 1$

$\rm \implies \: W = 0-( - 50.9 \times {10}^{ - 9})$

$\rm \implies \: W = 50.9 \times {10}^{ - 9} \: joule$

So, work done is 50.9 × 10^(-9) Joules.