Question

Two charges 10 × 10⁻⁹C and 20 × 10⁻⁹C are placed at a distance of 0.3 m apart. Find the potential and intensity at a point mid−way between them.

Answer 1

Given in the question :-
q₁ = 10 × 10⁻⁹ C
q₂ = 20 × 10⁻⁹ C
r = 0.3 m

For finding the potential :-
V = V₁ +V₂
V = (kq₁ /r₁ ) + (kq₂/r₂)
Here r₁ = r₂ = 0.3 m
V = k( q₁/r₁ + q₂/r₂)
V = 9 × 10⁹ [ ( 10 × 10⁻⁹ ) × (20 × 10⁻⁹)] / 0.3
V = 18 × 10⁻⁷ / 0.3
V = 6 × 10⁻⁶ V .


Now for the intensity ,We know the formula for finding the mid-way electric intensity  between the charges.. 
E = k[q₂ - q₁]/r²
E = [(9 × 10⁹) × [(20 × 10⁻⁹) - (10 × 10⁻⁹) ]/0.3²
E = (9 × 10⁹) × 10⁹(20 - 10) / 0.09
E = (9 × 10)/0.09 
E = 90/0.09
E = 1000 N/C




Hope it Helps.

Answer 2

✨HEY MATE ✨

HERE IS UR ANSWER ⤵️⤵️

Given in the question :-
q₁ = 10 × 10⁻⁹ C
q₂ = 20 × 10⁻⁹ C
r = 0.3 m

For finding the potential :-
V = V₁ +V₂
V = (kq₁ /r₁ ) + (kq₂/r₂)
Here r₁ = r₂ = 0.3 m
V = k( q₁/r₁ + q₂/r₂)
V = 9 × 10⁹ [ ( 10 × 10⁻⁹ ) × (20 × 10⁻⁹)] / 0.3
V = 18 × 10⁻⁷ / 0.3
V = 6 × 10⁻⁶ V .


Now for the intensity ,We know the formula for finding the mid-way electric intensity  between the charges.. 
E = k[q₂ - q₁]/r²

E = [(9 × 10⁹) × [(20 × 10⁻⁹) - (10 × 10⁻⁹) ]/0.3²

E = (9 × 10⁹) × 10⁹(20 - 10) / 0.09

E = (9 × 10)/0.09 

E = 90/0.09

E = 1000 N/C


$Hope \: that \: this \: answer \: will \: help \: u$