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Question: Two identical conducting spheres, fixed in space, attract each other with an electrostatic force of ...

Two identical conducting spheres, fixed in space, attract each other with an electrostatic force of 0.108 N when separated by 50 cm, centre-to-centre. A thin conducting wire then connects the spheres. When the wire is removed, the spheres repel each other with an electrostatic force of 0.036 N. What were the initial charges on the spheres?

Explanation

Solution

Use the given data and the formula for the electrostatic force between two charged spheres to find relation between the initial charges on both the spheres. Then equate the potential of the two when the wire is removed and find the new charge on the both spheres.

Formula used:
F=kq1q2r2F=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}
V=kqRV=\dfrac{kq}{R}

Complete step by step answer:
It is given that two identical spheres are attracting each other. Then this means that the charges on two spheres are of opposite signs. Let one of the spheres have charge q1{{q}_{1}} and the other sphere have a charge q2{{q}_{2}}.The electrostatic force between the two spheres is equal to F1=kq1q2r2{{F}_{1}}=\dfrac{k{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}, where k is a proportionality constant and r is the distance between the centres of the two spheres.
The value of k=9×109Nm2C−2k=9\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}}.

It is given that the two spheres are separated by 50 cm, centre-to-centre.
⇒d=50cm=0.5m\Rightarrow d=50cm=0.5m.
And it is said that force attraction between the two is equal to 0.108 N.
This means that F1=0.108N{{F}_{1}}=0.108N.
Substitute the known values in (i).
⇒0.108=(9×109)q1q2(0.5)2\Rightarrow 0.108=\dfrac{\left( 9\times {{10}^{9}} \right){{q}_{1}}{{q}_{2}}}{{{(0.5)}^{2}}}
⇒q1q2=(0.108)(0.5)2(9×109)=3×10−12\Rightarrow {{q}_{1}}{{q}_{2}}=\dfrac{(0.108){{(0.5)}^{2}}}{\left( 9\times {{10}^{9}} \right)}=3\times {{10}^{-12}} … (ii).
When the two spheres are connected by a conducting wire, charges will flow from a higher potential to a lower potential until they attain equal potential.
The potential of a conducting sphere is given as V=kqRV=\dfrac{kq}{R},
where q is the charge on the sphere and R is the radius of the sphere.

Since k is a constant and the radii of the spheres are equal, the charges on the both will be the same (potential is same).Let the new charge on both the spheres be q.However, the net charge of the system will be conserved. This means that the initial total charge and the new total charge are equal.
i.e. q1−q2=q+q{{q}_{1}}-{{q}_{2}}=q+q
⇒q=q1−q22\Rightarrow q=\dfrac{{{q}_{1}}-{{q}_{2}}}{2}.
It is said that after the wire is removed, the spheres repel each other by a force of 0.036 N.
⇒F2=0.036=kq.qr2\Rightarrow {{F}_{2}}=0.036=\dfrac{kq.q}{{{r}^{2}}}
Substitute the value of q, k and r.
⇒0.036=9×109(q1−q22)(q1−q22)(0.5)2\Rightarrow 0.036=\dfrac{9\times {{10}^{9}}\left( \dfrac{{{q}_{1}}-{{q}_{2}}}{2} \right)\left( \dfrac{{{q}_{1}}-{{q}_{2}}}{2} \right)}{{{(0.5)}^{2}}}
⇒(q1−q2)2=0.036×4×(0.5)29×109\Rightarrow {{\left( {{q}_{1}}-{{q}_{2}} \right)}^{2}}=\dfrac{0.036\times 4\times {{(0.5)}^{2}}}{9\times {{10}^{9}}}.
⇒(q1−q2)=0.036×4×(0.5)29×109=0.4×10−6C\Rightarrow \left( {{q}_{1}}-{{q}_{2}} \right)=\sqrt{\dfrac{0.036\times 4\times {{(0.5)}^{2}}}{9\times {{10}^{9}}}}=0.4\times {{10}^{-6}}C … (iii)

But from (i) we get that q2=3×10−12q1{{q}_{2}}=\dfrac{3\times {{10}^{-12}}}{{{q}_{1}}}.
Substitute this value in (iii).
⇒(q1−3×10−12q1)=0.4×10−6C\Rightarrow \left( {{q}_{1}}-\dfrac{3\times {{10}^{-12}}}{{{q}_{1}}} \right)=0.4\times {{10}^{-6}}C
⇒q12−3×10−12=0.4×10−6q1\Rightarrow q_{1}^{2}-3\times {{10}^{-12}}=0.4\times {{10}^{-6}}{{q}_{1}}
⇒q12−0.4×10−6q1−3×10−12=0\Rightarrow q_{1}^{2}-0.4\times {{10}^{-6}}{{q}_{1}}-3\times {{10}^{-12}}=0
By using quadratic formula, we get q1=−(0.4×10−6)±(0.4×10−6)2−4(1)(−3×10−12)2(1){{q}_{1}}=\dfrac{-(0.4\times {{10}^{-6}})\pm \sqrt{{{(0.4\times {{10}^{-6}})}^{2}}-4(1)(-3\times {{10}^{-12}})}}{2(1)}
⇒q1=−0.4×10−6±0.16×10−12+12×10−122\Rightarrow {{q}_{1}}=\dfrac{-0.4\times {{10}^{-6}}\pm \sqrt{0.16\times {{10}^{-1}}^{2}+12\times {{10}^{-12}}}}{2}
⇒q1=−0.4×10−6±10−612.162\Rightarrow {{q}_{1}}=\dfrac{-0.4\times {{10}^{-6}}\pm {{10}^{-6}}\sqrt{12.16}}{2}
⇒q1=−0.4×10−6±35.88×10−62\Rightarrow {{q}_{1}}=\dfrac{-0.4\times {{10}^{-6}}\pm 35.88\times {{10}^{-6}}}{2}

On solving for q1{{q}_{1}} using the quadratic formula, we get that q1=1.94×10−6C{{q}_{1}}=1.94\times {{10}^{-6}}C or q1=−1.54×10−6C{{q}_{1}}=-1.54\times {{10}^{-6}}C.
When q1=1.94×10−6C{{q}_{1}}=1.94\times {{10}^{-6}}C, q2=1.54×10−6C{{q}_{2}}=1.54\times {{10}^{-6}}C and when q1=−1.54×10−6C{{q}_{1}}=-1.54\times {{10}^{-6}}C, q2=−1.94×10−6C{{q}_{2}}=-1.94\times {{10}^{-6}}C
This means that if the charge on one sphere is q1=1.94×10−6C{{q}_{1}}=1.94\times {{10}^{-6}}C then the charge on the other sphere is q2=−1.54×10−6C{{q}_{2}}=-1.54\times {{10}^{-6}}C.
And if the charge on one sphere is q1=−1.54×10−6C{{q}_{1}}=-1.54\times {{10}^{-6}}C then the charge on the other sphere is q2=1.94×10−6C{{q}_{2}}=1.94\times {{10}^{-6}}C.

Note: If you do not know about the conservation of charge, then you can assume that the new charge on the two spheres to be q and equate the initial and final potential energy of the system. Since no external work is on the system, the potential energy of the system is conserved.