Question

27. The electric potential due to a point
charge at a distance of 20cm is 100V.
what is the strength of field at this point?
O
a. 400 V/m
O
b. 500 V/m.​

Answer 1

Answer:

Explanation:

REF.Image.

Electric field from distance = 100 N/m

R = Radius = 100m

when (r>R)  

So E=  

r  

2

 

Kθ

​  

 

100=  

20×20×10  

−4

 

Kθ

​  

 

Kθ=400×10  

−2

...(i)

and Electric field from centre distance is 3 cm.

(when r<R) E=  

R  

3

 

Kθr

​  

 from (i)equation

=  

10×10×10×10  

−6

 

400×10  

−2

×3×10  

−2

 

​  

 

=  

10  

−3

 

4×3×10  

−2

 

​  

=120V/m

E=120V/m

​