Question

33.Two electric charges of 9micro coloumb and 3micro coloumb are placed 0.16m apart in air.There are two points A and B on the line joining the two charges at distances of l) 0.04m from -3microcoloumb and inbetween the charges and (ll)0.08m from -3micro coloumb and outside the two charges.The potentials at A and B are

Answer 1

q1 = 9 μC        q2 =  - 3 μC            q2  is negative or positive ??
D = 0.16 meters
  
The electrostatic potential at a point P at a distance d from a charge q
         = V = q / ( 4 π ε₀ d )   = 9 * 10⁹ * q / d   volts 
If there are more than one , then algebraic sum of potentials is to be performed.

1)    PB = 0.04 m      PA = 0.12 m
         V = 9 * 10⁹ * [9 /0.12 - 3/0.04] * 10⁻⁶  volts = 0 volts

2)  PB = 0.08 m      PA = 0.24 m
 
         V = 9 * 10⁹ * [ 9 / 0.24 - 3/0.08 ] * 10⁻⁶  = 0 volts