Question

two point charges + 9 microcoulomb and minus 18 microcoulomb are separated by a distance of 2 m at what point along the line joining the charge is the net potential zero​

Answer 1

Answer:

1/3m from −2μC

Explanation:

Let potential is zero at a distance x from 4μc  

⇒  

x

k(4)

​

+  

(1−x)

k(−2)

​

=0

⇒  

x

4

​

=  

(1−x)

2

​

 

⇒2−2x=x

⇒2=3x

⇒x=  

3

2

​

 

∴ from the distance is 1−  

3

2

​

=  

3

1

​

 

potential is zero at  

3

1

​

m from −2μc.

Hence,

option (C) is correct answer.

Answer 2

Answer:

-2cm

Explanation:

$V_{net} = \frac{1}{4\pi \epsilon_0} \frac{9*10^{-6} }{2} + \frac{1}{4\pi \epsilon_0} \frac{-18*10^{-6} }{2-x}\\V_{net} =0 (given in the question)\\\frac{1}{4\pi \epsilon_0} [\frac{9*10^{-6} }{2} - \frac{18*10^{-6} }{2-x} ]= 0\\\\\frac{9}{2} - \frac{18}{2-x} =0\\On solving \\(2-x)9 =36\\18-9x = 36\\x= -2cm\\here -ve sign shows the opposite direction$