Question 787: Electrostatics & Capacitance

Question

Two point charges of $+3 \mu \mathrm{C}$ and $-3 \mu \mathrm{C}$ are placed at a certain distance apart from each other. Electric potential at a distance 60 cm from the mid-point of dipole on axial line is 150 V , then the value of dipole length is

Accepted Answer

Given:
Electric potential, $ V = 150 \, 	ext{V} $
Distance from the mid-point of the dipole on the axial line, $ r = 60 \, 	ext{cm} = 0.6 \, 	ext{m} $
The formula for the electric potential $ V $ on the axial line is:
$ V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p \cos 	heta}{r^2} $
For the axial line, $ 	heta = 0^\circ $.
Thus, the potential becomes:
$ V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{p \cdot 1}{r^2} $
Substitute the given values:
$ 150 = \frac{9 	imes 10^9 	imes p}{(0.6)^2} $
Solving for $ p $:
$ p = \frac{150 	imes 0.36}{9 	imes 10^9} $
So,
$ p = 6 	imes 10^{-9} \, 	ext{Cm} $
The dipole moment $ p = 2a \cdot q $.
Thus,
$ 2a = \frac{6 	imes 10^{-9}}{q} $
Given the charge $ q = 3 \, \mu	ext{C} = 3 	imes 10^{-6} \, 	ext{C} $:
$ 2a = \frac{6 	imes 10^{-9}}{3 	imes 10^{-6}} $
This simplifies to:
$ 2a = 2 	imes 10^{-3} \, 	ext{m} $
The length of the dipole, $ 2a $, is hence $ 2 	imes 10^{-3} \, 	ext{m} $.

Answer

$2 	imes 10^{-3} \mathrm{~m}$

Answer

$2 	imes 10^{-4} \mathrm{~m}$

Answer

$10^{-3} \mathrm{~m}$

Answer

$10^{-4} \mathrm{~m}$

Electrostatics & Capacitance

Two point charges of +3 \mu \mathrm{C} and -3 \mu \mathrm{C} are placed at a certain distance apart from each other. Electric potential at a distance 60 cm from the mid-point of dipole on axial line is 150 V , then the value of dipole length is

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