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

 

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 = 1.5 uC Magnitude of charge 1SuC
located at B, q2 2.5 μC

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

(a) Let Vi and Es are the electric potentials and electric field respectively at O.

Vi = Potential due to charge at A + Potential due to charge at 1 B

V1=q14πϵ0(d2)+44πϵ9(d2)=14πϵ0(d2)(q1+q2)

Where, Eo = Permittivity of free space

14πϵ0=9×109NC2m2V1=9×10×106(0.302)(2.5+1.5)=2.4×105V

E1= Electric field due to q2 Electric field due to q1=q24πϵ0(d2)2q14πϵ0(d2)2=9×109(0.302)2×106×(2.51.5)= 4 × 105 Vm-1

Therefore, the potential at mid-point is 2.4 x 105 V and the electric field at mid-point is 4 x 103 V m. The field is directed from the larger charge to the smaller charge.

(b) Consider a point Z such that normal distance 0Z = 10 cm =0.1 m, as shown in the following figure. V2 and E2 are the electric potentials and electric fields respectively at Z.

It can be observed from the figure that distance,

BZ=AZ=(0.1)2+(0.15)2=0.18m

V2 Electric potential due to A + Electric Potential due to B

=q14πϵ0(AZ)+q14πϵ0(BZ)=9×109×1060.18(1.5+2.5)=2×105V

Electric field due to q at Z,

EA=q14πϵ0(AZ)2=9×10×1.5×106(0.18)2=0.416×10V/m

Electric field due to q2 at Z,

EB=q24πϵ0(BZ)2=9×109×2.5×106(0.18)2=0.69×106Vm1

The resultant field intensity at Z,

E=EA2+EB2+2EAEBcos2θ

Where, 28 is the angle, AZ B

From the figure, we obtain

cosθ=0.100.18=59=0.5556θ=cos10.5556=56.252θ=112.5cos2θ=0.38E=(0.416×10)2×(0.69×10)2+2×0.416×0.69×102×(0.38)=6.6×104Vm4

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 x 105V and electric field is 6.6 X 105 Vm-1.