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Two Tiny Spheres Carrying Charges 1.5 μC and 2.5 μC Are Located 30 cm Apart. Find the Potential and Electric Field - Physics

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.

 
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Solution

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

Magnitude of charge located at A, q1 = 1.5 μC

Magnitude of charge located at B, q2 = 2.5 μC

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

(a) Let V1 and E1 are the electric potential and electric field respectively at O.

V1 = Potential due to charge at A + Potential due to charge at B

`V_1=q_1/(4piin_0(d/2))+q_2/(4piin_0(d/2))=1/(4piin_0(d/2))(q_1+q_2)`

Where,

0 = Permittivity of free space

`1/(4piin_0)=9xx10^9 NC^2 m^-2`

`therefore V_1=(9xx10^9xx10^-6)/((0.30/2))(2.5+1.5)=2.4xx10^5 V`

E1 = Electric field due to q2 − Electric field due to q1

`=q_2/(4piin_0(d/2)^2)-q_1/(4piin_0(d/2)^2)`

`(9xx10^9xx10^-6)/((0.30/2)^2)(2.5-1.5)=4xx10^5 Vm^-1`

Therefore, the potential at mid-point is 2.4 × 105 V and the electric field at mid-point is 4× 105 V m−1. The field is directed from the larger charge to the smaller charge.

(b) Consider a point Z such that normal distanceOZ = 10 cm = 0.1 m, as shown in the following figure.

V2 and Eare the electric potential and electric field respectively at Z.

It can be observed from the figure that distance,

`BZ=AZ=sqrt((0.1)^2+(0.15)^2)=0.18m`

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

`=q_1/(4piin_0(AZ))+q_1/(4piin_0(BZ))`

`=(9xx10^9xx10^-6)/0.18(1.5+2.5)`

Electric field due to q at Z,

`E_A=q_1/(4piin_0(AZ)^2)`

`=(9xx10^9xx1.5xx10^-6)/(0.18)^2`

`=0.416xx10^6 V/m`

Electric field due to q2 at Z,

`E_A=q_1/(4piin_0(bZ)^2)`

`=(9xx10^9xx2.5xx10^-6)/(0.18)^2`

`=0.69xx10^6 V/m`

The resultant field intensity at Z,

`E=sqrt(E_A^2+E_B^2+2E_AE_Bcos 2 theta)`

Where, 2θis the angle, ∠AZ B

From the figure, we obtain

`cos theta=0.10/0.18=5/9=0.5556`

`theta=cos^-1 0.5556=56.25`

`therefore 2 theta=-0.38`

`E=sqrt((0.416xx10^6)^2xx(0.69xx10^6)^2+2xx0.416xx0.69xx10^12xx(-0.38))`

`=6.6xx10^5 Vm^-1`

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

Concept: Potential Energy in an External Field - Potential Energy of a System of Two Charges in an External Field
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Q 14 | Page 88
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