$1.5\, \mu C$ और $2.5\, \mu C$ आवेश वाले दो सूक्ष्म गोले $30 \,cm$ दूर स्थित हैं।

$(a)$ दोनों आवेशों को मिलाने वाली रेखा के मध्य बिंदु पर, और

$(b)$ मध्य बिंदु से होकर जाने वाली रेखा के अभिलंब तल में मध्य बिंदु से $10\, cm$ दूर स्थित किसी बिंदु पर विभव और विध्यूत क्षेत्र ज्ञात कीजिए।

Two charges placed at points $A$ and $B$ are represented in the given figure. $O$ is the midpoint of the line joining the two charges.

Magnitude of charge located at $A, q_{1}=1.5\, \mu \,C$

Magnitude of charge located at $B, q_{2}=2.5\, \mu\, C$

Distance between the two charges, $d=30 \,cm =0.3 \,m$

$(a)$ Let $v_{1}$ and $E_{1}$ are the electric potential and electric field respectively at $0 .$

$V _{1}=$ Potential due to charge at $A +$ Potential due to charge at $B$

$V_{1}=\frac{q_{1}}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)}+\frac{q_{2}}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)}$$=\frac{1}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)}\left(q_{1}+q_{2}\right)$

Where, $\epsilon_{0}=$ Permittivity of free space

$\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} NC ^{2} m ^{-2}$

$\therefore V_{1}=\frac{9 \times 10^{9} \times 10^{-6}}{\left(\frac{0.30}{2}\right)}(2.5+1.5)$$=2.4 \times 10^{5} V$

$E _{1}=$ Electric field due to $q _{2}-$ Electric field due to $q _{1}=\frac{q_{2}}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)^{2}}-\frac{q_{1}}{4 \pi \epsilon_{0}\left(\frac{d}{2}\right)^{2}}$

$=\frac{9 \times 10^{9}}{\left(\frac{0.30}{2}\right)^{2}} \times 10^{6} \times(2.5-1.5)$

$=4 \times 10^{5}\, V \,m ^{-1}$

Therefore, the potential at mid-point is $2.4 \times 10^{5}\, V$ and the electric field at mid-point is $4 \times 10^{5} \,V\, m ^{-1} .$ The field is directed from the larger charge to the smaller charge.

$(b)$ Consider a point $z$ such that normal distance $OZ=10 \,cm =0.1\, m ,$ as shown in the following figure.

$V_{2}$ and $E_{2}$ are the electric potential and electric field respectively at $z$

It can be observed from the figure that distance, $B Z=A Z=\sqrt{(0.1)^{2}+(0.15)^{2}}=0.18\, m$

$V _{2}=$ Electric potential due to $A +$ Electric Potential due to $B$

$=\frac{q_{1}}{4 \pi \epsilon_{0}( AZ )}+\frac{q_{1}}{4 \pi \epsilon_{0}( BZ )}$

$=\frac{9 \times 10^{9} \times 10^{-6}}{0.18}(1.5+2.5)$

$=2 \times 10^{5}\, V$

Electric field due to $q$ at $z$ $E_{ A }=\frac{q_{1}}{4 \pi \epsilon_{0}( AZ )^{2}}$

$=\frac{9 \times 10^{9} \times 1.5 \times 10^{-6}}{(0.18)^{2}}$

$=0.416 \times 10^{6} \,V / m$

Electric field due to $q_{2}$ at $Z$ $E_{ n }=\frac{q_{2}}{4 \pi \epsilon_{0}( BZ )^{2}}$

$=\frac{9 \times 10^{9} \times 2.5 \times 10^{-6}}{(0.18)^{2}}$

$=0.69 \times 10^{6} \,V\,m ^{-1}$

The resultant field intensity at $z$

$E=\sqrt{E_{A}^{2}+E_{ B }^{2}+2 E_{ A } E_{ B } \cos 2 \theta}$

Where, $2 \theta$ is the angle, $\angle A Z B$

From the figure, we obtain

$\cos \theta=\frac{0.10}{0.18}=\frac{5}{9}=0.5556$

$\theta=\cos ^{-1} \theta \cdot 5556=56.25$

$\therefore 2 \theta=112.5^{\circ}$

$\cos 2 \theta=-0.38$

$E=\sqrt{\left(0.416 \times 10^{4}\right)^{2} \times\left(0.69 \times 10^{6}\right)^{2}+2 \times 0.416 \times 0.69 \times 10^{12} \times(-0.38)}$

$=6.6 \times 10^{5} \,V\,m ^{-1}$

Therefore, the potential at a point $10 \,cm$ (perpendicular to the mid-point) is $2.0 \times 10^{5} \,V$ and electric field is

$6.6 \times 10^{5} \,V\, m ^{-1}$