Question

ABCD is a square of side 2m. Point charges of 5nC, 10nC and -5nC are

placed at corners A, B, C respectively. Calculate the work done in

transferring a charge of 5nC from D to the point of intersection of diagonals. please answer this question with steps..​

Answer 1

Answer:

the work done in  transferring a charge of 5nC from D to the point of intersection of diagonals is

$W = 1.59 \times 10^{-7} J$

Explanation:

Initial potential energy of the charge placed at the position D is given as

$U_i = \frac{(9\times 10^9)(5\times 10^{-9})(5\times 10^{-9}}{2} + \frac{(9\times 10^9)(10\times 10^{-9})(5\times 10^{-9}}{2\sqrt2} + \frac{(9\times 10^9)(-5\times 10^{-9})(5\times 10^{-9}}{2}$

$U_i = 1.125 \times 10^{-7} + 1.59 \times 10^{-7} - 1.125\times 10^{-7}$

$U_i = 1.59 \times 10^{-7} J$

Now the final potential energy of the system is given as

$U_f = \frac{(9\times 10^9)(5\times 10^{-9})(5\times 10^{-9}}{\sqrt2} + \frac{(9\times 10^9)(10\times 10^{-9})(5\times 10^{-9}}{\sqrt2} + \frac{(9\times 10^9)(-5\times 10^{-9})(5\times 10^{-9}}{\sqrt2}$

$U_f = 1.59 \times 10^{-7} + 3.18 \times 10^{-7} - 1.59\times 10^{-7}$

$U_f = 3.18 \times 10^{-7} J$

So work done in moving the charge is given as

$W = U_f - U_i$

$W = (3.18 - 1.59)\times 10^{-7} J$

$W = 1.59 \times 10^{-7} J$

#Learn

Topic : Electrostatic potential energy

https://brainly.in/question/9352332