Question 789: Electrostatics & Capacitance

Question

The net electric force on a charge of $$+3 \mu \mathrm{C}$$ at the mid-point on the line joining two charges of magnitude $$+2 \mu \mathrm{C}$$ and $$-2 \mu \mathrm{C}$$ separated by the distance of $$6 \mathrm{~mm}$$, is

Accepted Answer

Let's carefully solve this step by step using the
CoulombÃ¢ÂÂs law
as given in the NCERT syllabus.
Given data:
Charge $ q_1 = +2~\mu C = +2 	imes 10^{-6}~C $
Charge $ q_2 = -2~\mu C = -2 	imes 10^{-6}~C $
Test charge at midpoint $ q_0 = +3~\mu C = +3 	imes 10^{-6}~C $
Distance between $ q_1 $ and $ q_2 $ = $ 6~mm = 6 	imes 10^{-3}~m $
Hence,
distance of the midpoint from each charge =
$ r = \frac{6 	imes 10^{-3}}{2} = 3 	imes 10^{-3}~m $
Step 1: Find the force on $ q_0 $ due to each charge
By
CoulombÃ¢ÂÂs law
,
$ F = k \frac{|q_1 q_0|}{r^2} $
where $ k = 9 	imes 10^9~N\,m^2/C^2 $.
So, the magnitude of each force on $ q_0 $ is
$ F = 9 	imes 10^9 	imes \frac{(2 	imes 10^{-6})(3 	imes 10^{-6})}{(3 	imes 10^{-3})^2} $
$ F = 9 	imes 10^9 	imes \frac{6 	imes 10^{-12}}{9 	imes 10^{-6}} $
$ F = 9 	imes 10^9 	imes 6.67 	imes 10^{-7} = 6000~N $
Step 2: Determine the directions
The charge $ +2~\mu C $
repels
the test charge $ +3~\mu C $.
So the force due to $ +2~\mu C $ is
away from it
(say, toward right).
The charge $ -2~\mu C $
attracts
the $ +3~\mu C $.
So the force due to $ -2~\mu C $ is
toward it
, which is
also toward right
(if $ -2~\mu C $ is on the right side).
Thus,
both forces act in the same direction
.
Step 3: Net force
$ F_{	ext{net}} = F_1 + F_2 = 6000 + 6000 = 12000~N $
Step 4: Final answer
$ \boxed{F_{	ext{net}} = 12000~N} $
Correct option: (C) 12000 N

Answer

6000 N

Answer

500 N

Answer

12000 N

Answer

zero

Electrostatics & Capacitance

The net electric force on a charge of $+3 \mu \mathrm{C} at the mid-point on the line joining two charges of magnitude +2 \mu \mathrm{C} and -2 \mu \mathrm{C} separated by the distance of 6 \mathrm{~mm}$, is

Choose the correct answer: