# 2.14  Two tiny spheres carrying charges $1.5 \mu C$ and $2.5 \mu C$ are located 30 cm apart. Find the potential and electric field:(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

The distance of the point from both the charges :

$d=\sqrt{0.1^2+0.15^2}=0.18m$

Hence,

Electric potential:

$V=\frac{kq_1}{d}+\frac{kq_2}{d}=\frac{k}{0.18}(1.5+2.5)*10^{-6}=2*10^5V$

Electric field due to q1

$E_1=\frac{kq_1}{d^2}=\frac{k1.5\mu C}{0.18^2m^2}=0.416*10^6V/m$

Electric field due to q2

$E_2=\frac{kq_2}{d^2}=\frac{k2.5\mu C}{0.18^2m^2}=0.69*10^6V/m$

NOW,

Resultant Electric field :

$E=\sqrt{E_1^2+E_2^2+2E_1E_2cos\theta}$

Where $\theta$ is the angle between both electric field directions

Here,

$cos\frac{\theta}{2}=\frac{0.10}{0.18}=\frac{5}{9}$

$\frac{\theta}{2}=56.25$

${\theta}=2*56.25=112.5$

Hence

$E=\sqrt{(0.416*10^6)^2+(0.69*10^6)^2+2(0.416*10^6)(0.69*10^6)cos112.5}$

$E=6.6*10^5V/m$

Exams
Articles
Questions