Question

What is the magnitude of the point charge chosen so that electric field 20cm away from it has a magnitude of 18*10N/C

Answer 1

Answer:

let magnitude of charge be Q.  

so,magnitude of electric field (E) at a distance r=0.5m away in given by:-  

E  

r  

2

 

KQ

​  

 , which gives:  

Q=  

k

Er  

2

 

​  

=  

9×10  

9

 

2×(0.5)  

2

 

​  

≃5.56×10  

−11

c

approximately Q=55.6pc

Explanation: