Question 17: Electrostats & Capacitance

Question

Two short dipoles $(A, B)$, $A$ having charges $\pm 2\mu	ext{C}$ and length $1	ext{ cm}$ and $B$ having charges $\pm 4\mu	ext{C}$ and length $1	ext{ cm}$ are placed with their centres $80	ext{ cm}$ apart as shown in the figure. The electric field at a point $P$, equi-distant from the centres of both dipoles is ______ $	ext{N/C}$.

Accepted Answer

wo short dipoles $(A, B)$ are placed with their centres $80	ext{ cm}$ apart.Dipole A: Charges $q_A = \pm 2\mu	ext{C}$, Length $l_A = 1	ext{ cm}$Dipole B: Charges $q_B = \pm 4\mu	ext{C}$, Length $l_B = 1	ext{ cm}$Distance to Point P: $r = 40	ext{ cm} = 0.4	ext{ m}$ (midpoint)$\$Step 1: Calculate Dipole Moments$\$ $p_A = q_A 	imes l_A = (2 	imes 10^{-6}	ext{ C}) 	imes (0.01	ext{ m}) = 2 	imes 10^{-8}	ext{ C}\cdot	ext{m}$$$$p_B = q_B 	imes l_B = (4 	imes 10^{-6}	ext{ C}) 	imes (0.01	ext{ m}) = 4 	imes 10^{-8}	ext{ C}\cdot	ext{m}$$ $\$Step 2: Determine Field Directions at PField due to A ($E_A$): Point $P$ lies on the axial line of $A$.$$E_A = \frac{1}{4\pi\epsilon_0} \frac{2p_A}{r^3} = \frac{(9 	imes 10^9) 	imes (2 	imes 2 	imes 10^{-8})}{(0.4)^3}$$$$E_A = \frac{360}{0.064} = 5.625 	imes 10^3	ext{ N/C}$$Field due to B ($E_B$): Point $P$ lies on the equatorial line of $B$ (perpendicular).$$E_B = \frac{1}{4\pi\epsilon_0} \frac{p_B}{r^3} = \frac{(9 	imes 10^9) 	imes (4 	imes 10^{-8})}{(0.4)^3}$$$$E_B = \frac{360}{0.064} = 5.625 	imes 10^3	ext{ N/C}$$. $\$Step 3: Resultant FieldSince the fields are perpendicular ($\vec{E}_A \perp \vec{E}_B$):$$E_{	ext{net}} = \sqrt{E_A^2 + E_B^2} = E_A \sqrt{2}$$ $$E_{	ext{net}} = 5.625 \sqrt{2} 	imes 10^3	ext{ N/C} = \frac{9}{16} \sqrt{2} 	imes 10^4	ext{ N/C}$$ $$\$Correct Option: (B)

Answer

A. $\frac{9}{16} \sqrt{2} 	imes 10^5$

Answer

B. $\frac{9}{16} \sqrt{2} 	imes 10^4$

Answer

C. $9\sqrt{2} 	imes 10^4$

Answer

D. $4.5\sqrt{2} 	imes 10^4$

Electrostats & Capacitance

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