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.

Answers (1)

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

 

 

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