CBSE Class 12 · 2019 · ElectrostaticsLong Answer

(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
(b) Two identical point charges,

q

each, are kept 2 m apart in air. A third point charge

Q

of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of

Q

Solution

(a) Let P be an equatorial point at distance

r

; each charge is at

r^{2} + a^{2}

E_{+} = E_{-} = \frac{1}{4 π ε _{0}} \frac{q}{r ^{2} + a ^{2}}

Perpendicular components cancel; axial components add, with

cos θ = \frac{a}{r ^{2} + a ^{2}}

E = \frac{1}{4 π ε _{0}} \frac{p}{( r ^{2} + a ^{2} ) ^{3/2}}

anti-parallel to

p

; for

r ≫ a

E = \frac{1}{4 π ε _{0}} \frac{p}{r ^{3}}

.
(b) By symmetry

Q

sits at the mid-point, 1 m from each

q

.
For an outer charge to be in equilibrium, attraction by

Q

equals repulsion by the other

q

\frac{1}{4 π ε _{0}} \frac{q ∣ Q ∣}{( 1 ) ^{2}} = \frac{1}{4 π ε _{0}} \frac{q ^{2}}{( 2 ) ^{2}}

∣ Q ∣ = \frac{q}{4}

The force must be attractive, so

Q

is negative:

Q = - \frac{q}{4} at the mid-point

CBSE Class 12 Board · 2019 · Subjective