Question

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 mid-point in a plane normal to the line and passing through the mid-point.

Solution

Given,
Charge on 1st sphere, q₁=

1.5 μC

Charge on 2nd sphere, q₂=

2.5 μC

Distance between the two charges, r= 30 cm

(a) Potential at the mid-point of the line joining the two charges is

V = \frac{1}{4 {πε}_{0}} [\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}] = 9 \times 10^{9} [\frac{1.5 \times 10^{- 6}}{0.15} + \frac{2.5 \times 10^{- 6}}{0.15}] V = 9 \times 10^{9} \times \frac{10^{- 6}}{0.15} [4] = 240000

\Rightarrow

V = 2.4 \times 10^{5} V

Electric field at the mid point O due to charge at A

E = \frac{1}{4 {πε}_{0}} \frac{q_{1}}{{r_{1}}^{2}} = 9 \times 10^{9} \times \frac{1.5 \times 10^{- 6}}{(0.15)^{2}} = 6 \times 10^{5} {Vm}^{- 1} along OB

Electric field at the mid-point O due to charge at B

E = \frac{1}{4 {πε}_{0}} \frac{q_{2}}{{r_{2}}^{2}} = 9 \times 10^{9} \times \frac{2.5 \times 10^{- 6}}{(0.15)^{2}} = 10 \times 10^{5} {Vm}^{- 1} along OA

Thus, the total electric field at the mid-point O is

E = 10 \times 10^{5} - 6 \times 10^{5} = 4 \times 10^{5} {Vm}^{- 1} (along BA)

(b) Potential at the point C due to the two charges is

V = \frac{1}{4 {πε}_{0}} [\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}] w h e r e, r_{1} i s t h e d i s t a n c e f r o m A t o C a n d, r_{2} i s t h e d i s t a n c e f r o m B t o C r_{1} = r_{2} = \sqrt{(15)^{2} + (10)^{2}} = \sqrt{325} c m = \sqrt{325} \times 10^{- 2} m = 9 \times 10^{9} [\frac{1.5 \times 10^{- 6}}{\sqrt{325} \times 10^{- 2}} + \frac{2.5 \times 10^{- 6}}{\sqrt{325} \times 10^{- 2}}] V = \frac{9 \times 10^{9} \times 10^{- 6}}{10^{- 2}} . \frac{4.0}{\sqrt{325}} V = \frac{9 \times 4}{18.02} \times 10^{5} V = 2 \times 10^{5} V

Electric field at C due to charge at A

Electric field at C due to charge at B

E_{2} = \frac{1}{4 {πε}_{0}} \frac{q_{2}}{{r_{2}}^{2}} = 9 \times 10^{9} \times \frac{2.5 \times 10^{- 6}}{325 \times 10^{- 4}} = 6.92 \times 10^{5} {Vm}^{- 1}

If the angle between E₁ and E₂ be θ, then
Using law of trignometry,
tan

θ

\frac{side opposite to angle}{perpendicular}

\tan \frac{θ}{2} = \frac{0.15}{0.10} = 1.5

θ / 2 = 56.3 ° or, θ = 112.6 °

Thus, magnitude of resultant field at C is

For Daily Free Study Material Join wiredfaculty