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

Explanation

(a) Given, charges on the two spheres, `q_(1)=1.5muC, q_(2)=2.5muC`

Electric potential at the midpoint of the line joining the two spheres can be calculated as the sum of the electric potential due to individual charges:
`V=(1)/(4pi epsilon_(0)) ((q_(1))/(r_(1))+(q_(2))/(r_(2)))`
`=9xx10^(9)((1.5xx10^(-6))/(0.15)+(2.5xx10^(-6))/(0.15))`
`rArr V=2.4xx10^(5)V`
Net electric field at point O is given by:
`vecE=vecE_(2)-vecE_(1)`
The magnitude of electric field at the midpoint of the line joining the two spheres can be calculated as:
`rArr E=((q_(2)-q_(1)))/(4pi epsilon_(0)r^(2))=(9xx10^(9)xx(2.5xx10^(-6)-1.5xx10^(-6)))/((15xx10^(-2))^(2))`
`=4.0xx10^(5)Vm^(-1)`
The direction of electric field at point O is along BA.
(b)
From the given figure, `PA=sqrt(0.15^(2)+0.1^(2))=0.18m`
Potential at point P due to both the charges can be calculated as:
`V=(1)/(4pi epsilon_(0)) [(q_(1))/(AP)+(q_(2))/(BP)]=9xx10^(9)((q_(1))/(AP)+(q_(2))/(BP))`
`V=(9xx10^(9))/(0.18)(1.5xx10^(-6)+2.5xx10^(-6))=2xx10^(5)V`
The electric field at point P due to electric charges `q_(1) and q_(2)` are given as follows:
`E_(1)=(q_(1))/(4pi epsilon_(0)(PA)^(2))`
`=((9xx10^(9))xx(1.5xx10^(-6)))/((0.18)^(2))`
`=4.16xx10^(5)Vm^(-1)`
`E_(2)=(q_(2))/(4pi epsilon_(0)(PB)^(2))`
`=(9xx10^(9))/((0.18)^(2))xx(2.5xx10^(-6))=6.94xx10^(5)Vm^(-1)`
The magnitude of net electric field at point P due to both the charges can be calculated as:
`E=sqrt(E_(1)^(2)+E_(2)^(2)+2E_(1)E_(2)cos alpha)`
From the figure, we can understand
`alpha=180^(@)-(theta+theta)=180^(@)-2theta`
`cos theta=(0.15)/(0.18)=(5)/(6) rArr theta=33.56^(@)`
`therefore 2theta=67.12^(@)`
`alpha=180^(@)-2theta=112.88^(@)`
`E=sqrt((4.16xx10^(5))^(2)+(6.94xx10^(5))^(2)+2xx4.16xx10^(5)xx6.94xx10^(5)cos (112.88^(@)))`
`=10^(5)xx(sqrt(17.30+48.16-22.44))`
`rArr E=6.5xx10^(5)N//C`
Let `beta` be the angle between the resultant electric field intenstiy `vecE` and electric field intensity `vecE_(1)`
`tan beta=(E_(2)sin alpha)/(E_(1)+E_(2)cos alpha)`
`rArr tan beta=(6.94sin alpha)/(4.16+6.94cos alpha)`
`=(6.94sin(112.88^(@)))/(4.16+6.94 cos (112.88^(@)))=(6.39)/(1.46)`
`=4.37`
`beta=77.12^(@)`
Using geometry we can find the angle between net electric field and AB:

The net electric field intensity is inclined at an angle
`delta=180^(@)-(beta+theta)` with side AB
`delta =69.32^(@)`