Question

two charges are placed on the x-axis + 3 microcoulomb at X equal to zero and a charge of + 5 microcoulomb at X equal to 0.4 metre where must be a third charge placed if the force it experiences to be zero? Please show me how to calculate the answer with full steps

Test on Monday. Please explain faster

Answer 1

Force between 2 charges 'a' and 'b' seperated by a distance 'r' is given as,
F=k×[(a×b)/r²]
Where 'k' is constant.

Now let's consider the third charge 'Q' to be placed at 'R' distance from +3micro coulomb. [Only a assumption]

To be in equilibrium,
Force between +3 and Q = Force between Q and +5
=>k×[(3×Q)/(R-0)²] = k×[(5×Q)/(0.4-R)²]
=> 3Q/R² = 5Q/(0.16-0.8R+R²)
=>3/R² = 5/(0.16-0.8R+R²)
=>0.48-2.4R+3R² = 5R²
=>2R²+2.4R-0.48 = 0

The above given is a quadratic and on solving we get that,
R = (-3+√15)/5 = 0.175
Or
R=(-3-√15)/5 = -1.37

Since the second R (-1.37) can not be taken because we are taken a charge between 0 and 0.4 ,it can not be a negative value.
So, R=0.175

Hence we need to place the third charge at a distance 0.175 from origin on X-axis

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