Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Explanation

Two charges 1.5 `mu`C and 2.5 `mu`C are placed at point A and B respectively. O is the mid point of the line joining two charges. The distance between these charges is 30 cm.

Electrical potential at point O.
Electrical potential at point O
`V_(O)= (kq_(1))/(r_(1))+(kq_(2))/(r_(2))`
`=k[(1.5xx10^(-6))/(0.15)+(2.5xx10^(-6))/(0.15)]`
`=9xx10^(9)[(4xx10^(-6))/(0.15)]`
`:. V_(O)= 2.4xx10^(5)V`
Electric field at point O .
`E_(O)=E_(2)-E_(1) = (kq_(2))/(r_(2)^(2))-(kq_(1))/(r_(1)^(2))`
`:. E_(O)=k[(q^(2))/(qr_(2)^(2))-(q_(1))/(r_(1)^(2))]=9xx10^(9)`
`[(2.5xx10^(-6))/((0.15)^(2))-(1.5xx10^(-6))/((0.15)^(2))]`
`:. E_(O)= 9xx10^(9)xx(1.0xx10^(-6))/(0.0225)`
`:. E_(O)=4xx10^(5) Vm^(-1) rarr ` From `2.5 mu ` C to `1.5 muC`
(b) Let P is on the perpendicular bisector from mid point of AB at a distance 10 cm .

AP=BP = `sqrt(15^(2)+10^(2))`
`= sqrt(325)`
= 18 cm = 0.18 m
Electric potential at point P
`V_(p)=k[(q_(1))/(AP)+(q_(2))/(BP)]`
`= 9xx10^(9)[(1.5xx10^(-6))/(0.18)+(2.5xx10^(-6))/(0.15)]`
`= 9xx10^(9)xx(4xx10^(-6))/(0.18)`
`:. V_(P)= 2xx10^(5)` V
Electric field at P due to `q_(1)`
`E_(1)= (kq_(1))/((AP)^(2))=(9xx10^(9)xx1.5xx10^(-6))/((0.18)^(2))`
`:. E_(1)=0.42xx10^(6) Vm^(-1)` along AP
Electric field at P due to `q_(2)`
`E_(2) = (kq^(2))/((BP)^(2))= = (9xx10^(9)xx2.5xx10^(-6))/((0.18)^(2))`
`:. E_(2) = 0.69xx10^(6) Vm^(-1) ` along BP
Now `angleAPO=angleBPO= (theta)/(2)`
and in `DeltaAOP`
`tan""(theta)/(2) = 56.3^(@)`
`:. theta= 112.6^(@)`
Resultant electric field ,
`E= sqrt(E_(1)^(2)+E_(2)^(2)+2E_(1)E_(2)costheta)`
`=sqrt((0.42xx10^(6))^(2)+(0.69xx10^(6))^(2)+2xx0.42xx0.69xx10^(12)xxcos112.6^(@))`
`=[sqrt(0.1764+0.4761+0.2898xxcos112.6^(@))]xx10^(6)`
`= sqrt(0.6525+0.2898xx(-0.3843))xx10^(6)`
`:. E = sqrt(0.6525-0.2224)xx10^(6)`
`:. E = sqrt(0.4301)xx10^(6)`
`E= 65559xx10^(6)`
`:. E= 6.6xx10^(6)Vm^(-1)`
Suppose `alpha` is the angle made by E with `E_(1)`
`:. tan alpha = (E_(2)sin theta)/(E_(1)+E_(2)cos theta)`
`= (0.69xx10^(6)sin 112.6^(@))/(0.42xx10^(6)+0.69xx10^(6)cos112.6^(@))`
`[ because sin 112.6^(@) = sin 67.4^(@)`]
`=(0.69xx10^(6)xx0.9239)/(0.42xx10^(6)+0.69xx10^(6)xx(-0.3843))`
`= (0.6375xx10^(6))/(0.1548xx10^(6))`
`:. tan alpha= 4.1182`
`:. alpha= tan^(-1) 4.1182 `
`:. alpha=76.4^(@)`
Suppose `beta` is the angle made by E with BA
`:. angle PCO =beta`
`:.` In `DeltaPOC`
`beta+ alpha-(theta)/(2)= 90^(@)`
`:. beta= 90^(@)+(theta)/(2) -alpha`
`:. beta= 90^(@) +(112.6^(@))/(2) - 76.3^(@)`
`:. beta = 90^(@) + 56.3^(@) - 76.3^(@)`
`beta = 70^(@)`
Hence, resultant electric field make an angle `70^(@)` to the line joining between charges `2.5 muC` to `1.5 mu`C.