Two particles A and B , possessing opposite charges of +2.0 × 10^ -6 C and -2.0 × 10^ -6 C , are positioned at a separation of 1.0 cm . Evaluate the electric field at a point located on the perpendicular bisector of the

Correct answer: D. 180 N C^ -1 Given charges q = 2.0 × 10^ -6 C and separation 2a = 1.0 cm = 10^ -2 m . Dipole moment p = q × 2a = 2.0 × 10^ -6 × 10^ -2 = 2.0 × 10^ -8 C m . Distance from the centre r = 1.0 m . Since r 2a , the electric field at a point on the perpendicular bisector (equatorial plane) is given by E = (1)/(4π _0) (p)/(r^3) . Substituting the values: E = 9 × 10^9 × 2.0 × 10^ -8 (1.0)^3 E = 180 N C^ -1 .

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: Electric Field and Potential

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Electric Field and Potential

Two particles A and B , possessing opposite charges of +2.0 × 10^...

Two particles

A

and

B

, possessing opposite charges of

+2.0 \times 10^{-6} \text{ C}

and

-2.0 \times 10^{-6} \text{ C}

, are positioned at a separation of

1.0 \text{ cm}

. Evaluate the electric field at a point located on the perpendicular bisector of the dipole,

1.0 \text{ m}

away from the centre.

Options

A $90 \text{ N C}^{-1}$
B $720 \text{ N C}^{-1}$
C $360 \text{ N C}^{-1}$
D $180 \text{ N C}^{-1}$

Correct answer

D. $180 \text{ N C}^{-1}$

Step-by-step solution

Given charges

q = 2.0 \times 10^{-6} \text{ C}

and separation

2a = 1.0 \text{ cm} = 10^{-2} \text{ m}

.

Dipole moment

p = q \times 2a = 2.0 \times 10^{-6} \times 10^{-2} = 2.0 \times 10^{-8} \text{ C m}

.

Distance from the centre

r = 1.0 \text{ m}

.

Since

r \gg 2a

, the electric field at a point on the perpendicular bisector (equatorial plane) is given by

E = \dfrac{1}{4\pi \epsilon_0} \dfrac{p}{r^3}

.

Substituting the values:

E = \dfrac{9 \times 10^9 \times 2.0 \times 10^{-8}}{(1.0)^3}

E = 180 \text{ N C}^{-1}

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