Question

V4 m (vii) Two positive charges of 20 C and 8 hCare 20 cm apart then work done in bringing 5 cm closer is ​

Answer 1

Answer:

q  

1

​

=20×10  

−6

c , q  

2

​

=8×10  

−6

c

So, the primary distance r  

1

​

=20cm=0.2m

Final distance (20-5) = 15 = 0.15m

Now the work done =  

R

kq  

1

​

q  

2

​

 

​

 

=  

(0.15−0.2)

9×10  

9

×20×10  

−6

×8×10  

−6

 

​

×0.2

=5.7Joule

​

 

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