Question

Two positive point charges of 9uC and 6uC are
10 cm apart. The work done in bringing them to a
separation of 4 cm is
(1) 5.25 J
(2) 6.29 J
(3) 7.29 J
(4) 8.25 J​

Answer 1

Explanation:

First chargeq_{1}=12\times10^{-6}\ Cq

1

=12×10

−6

C

Second chargeq_{2}=8\times10^{-6}\ Cq

2

=8×10

−6

C

The distance r_{1}= 10\ cm=0.1\ mr

1

=10 cm=0.1 m

The distance r_{2}= 6\ cm=0.6\ mr

2

=6 cm=0.6 m

The work done will be

W= kq_{1}q_{2}(\dfrac{1}{r_{2}}-\dfrac{1}{r_{1}})W=kq

1

q

2

(

r

2

1

−

r

1

1

)

W=9\times10^{9}\times12\times10^{-6}\times8\times10^{-6}\times(\dfrac{1}{0.06}-\dfrac{1}{0.1})W=9×10

9

×12×10

−6

×8×10

−6

×(

0.06

1

−

0.1

1

)

W= 5.76\ JW=5.76 J

Hence, The work done will be 5.76 J.