# 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, - Physics

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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:

(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.

#### 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 potentials 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) = 9 xx 10^9  "N C"^2  "m"^-2

∴ "V"_1 = (9 xx 10^9 xx 10^-6)/((0.30/2))(2.5 + 1.5)

= 2.4 xx 10^5  "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)

= (9 xx 10^9)/((0.30/2)^2) xx 10^6 xx (2.5 - 1.5)

= 4 xx 10^5  "V m"^-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 distance OZ = 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) = 0.18 m

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

= "q"_1/(4piin_0("AZ")) + "q"_2/(4piin_0("BZ"))

= (9 xx 10^9 xx 10^-6)/0.18(1.5 + 2.5)

Electric field due to q at Z,

"E"_"A" = "q"_1/(4piin_0("AZ")^2)

= (9 xx 10^9 xx 1.5 xx 10^-6)/(0.18)^2

= 0.416 xx 10^6 "V"/"m"

Electric field due to q2 at Z,

E_"B" = "q"_2/(4piin_0("BZ")^2)

= (9 xx 10^9 xx 2.5 xx 10^-6)/(0.18)^2

= 0.69 xx 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, 2θ is the angle, ∠AZB

From the figure, we obtain

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

theta = cos^-1 0.5556 = 56.25

∴ 2theta = 112.5°

cos2 theta = -0.38

"E" = sqrt((0.416 xx 10^6)^2 xx (0.69 xx 10^6)^2 + 2 xx 0.416 xx 0.69 xx 10^12 xx (-0.38))

= 6.6 xx 10^5  "V m"^-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 Physics Part 1 and 2 Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.14 | Page 87
NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 14 | Page 88
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