Question

Ob Two
Two point charges are separated by a distance of 10 cm. Charge on point A =-9 C
and charge on point B = 4uC. k= 9 x 109 Nm²c-?. 1 °C = 10-6C. What is the change
in electric
potential
energy of charge on point B if accelerated to point A?
10 cm
A
B​

Answer 1

Answer:

Formula of Potential energy  

Electrostatic potential energy is given by U between two charges placed at a separtion r is given by  

⇒  U=  

r

kq  

1

​

q  

2

​

 

​

 

Step 2: Calculations

Putting values, k=9×10  

9

 

C  

2

 

N.m  

2

 

​

,  

q  

2

​

=q  

1

​

=1μC=1×10  

−6

C,

r=1m

⇒U

=  

1

9×10  

9

×1×10  

−6

×1×10  

−6

 

​

=9×10  

−3

J

Explanation: