This page focuses on the detailed Electrostatic Potential and Capacitance question answers for Class 12 Physics Electrostatic Potential and Capacitance, addressing the 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 midpoint in a plane normal to the line and passing through the mid-point.'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.

Question 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.

Answer

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, q_{1} = 1.5 μC

Magnitude of charge located at B, q_{2} = 2.5 μC

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

(a) Let *V _{1}* and

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

Where, ∈0 = Permittivity of free space

*E _{1}* = Electric field due to q

Therefore, the potential at mid-point is 2.4 × 10^{5} V and the electric field at mid-point is 4 × 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 distanceOZ = 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,

V_{2}= Electric potential due to A + Electric Potential due to B

Electric field due to q at Z,

Electric field due to q_{2} at Z,

The resultant field intensity at Z,

Where, 2θis the angle, ∠AZ B

From the figure, we obtain

Therefore, the potential at a point 10 cm (perpendicular to the mid-point) is 2.0 × 10^{5} V and electric field is 6.6 ×10^{5} V m ^{-1}

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