JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Electrostatics

An electric dipole is placed on

x

- axis in proximity to a line charge of linear density

3.0 \times 1 0^{- 6} C / m

. Line charge is placed on z -axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4 N is exerted on the dipole, find out the amount of positive charge of the dipole.

(A)0.485 mC
(B) $8.8 μ C$
(C)815.1 nC
(D) $4.44 μ C$ ✓ Correct

Answer: (D)

Solution

The electric field due to an infinite line charge at distance

r

E = \frac{λ}{2 π ϵ _{0} r}

. The force on the dipole is

F = q (E_{+} - E_{-})

where

E_{+}

and

E_{-}

are fields at positive and negative charge locations.

F = q λ / (2 π ϵ_{0}) \times (1/ r_{+} - 1/ r_{-})

where

r_{+} = 0.01

m and

r_{-} = 0.012

m. Substituting:

4 = q \times \frac{3.0 \times 1 0 ^{- 6}}{2 π \times 8.85 \times 1 0 ^{- 12}} \times (\frac{1}{0.01} - \frac{1}{0.012})

. Solving:

4 = q \times 5.4 \times 1 0^{4} \times 16.67

, giving

q = 4.44 \times 1 0^{- 6}

= 4.44 μ

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Electrostatics

Solution

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