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Question: Two identical conducting spheres having unequal opposite charges attract each other with a force of ...

Two identical conducting spheres having unequal opposite charges attract each other with a force of 3.15 N when separated by 0.2 m. The sphere experiences a force of repulsion of 0.625 N when they are made to touch for moment and then placed at a distance 0.3 m apart. Find the initial charge on each sphere.

Answer

q1 = +7.0 × 10^{-6} C, q2 = -2.0 × 10^{-6} C

Explanation

Solution

Solution:

Let the charges be q1q_1 and q2q_2 (with q1>0q_1>0 and q2<0q_2<0). We are given two conditions:

  1. Before Contact
    The spheres, when separated by r=0.2 mr=0.2\,\text{m}, attract with force

    F=k∣q1q2∣r2=3.15 N.F = k\frac{|q_1q_2|}{r^2} = 3.15\ \text{N}.

    Using k=9×109 N m2/C2k = 9\times10^9 \, \text{N m}^2/\text{C}^2, we get

    ∣q1q2∣=Fr2k=3.15×(0.2)29×109=3.15×0.049×109=0.1269×109=1.4×10−11 C2.|q_1q_2| = \frac{F r^2}{k} = \frac{3.15 \times (0.2)^2}{9\times10^9} = \frac{3.15 \times 0.04}{9\times10^9} = \frac{0.126}{9\times10^9} = 1.4\times10^{-11}\, \text{C}^2.

  2. After Contact
    When touched, the charges redistribute equally so that each gets

    q1+q22.\frac{q_1+q_2}{2}.

    They are then repelled with force F′=0.625 NF' = 0.625\,\text{N} at r′=0.3 mr'=0.3\,\text{m}:

    F′=k(q1+q22)2(0.3)2.F' = k \frac{\Bigl(\frac{q_1+q_2}{2}\Bigr)^2}{(0.3)^2}.

    Substituting values:

    0.625=9×109⋅(q1+q22)20.09.0.625 = 9\times10^9 \cdot \frac{\left(\frac{q_1+q_2}{2}\right)^2}{0.09}.

    Rearranging,

    9×1090.09⋅(q1+q2)24=0.625.\frac{9\times10^9}{0.09} \cdot \frac{(q_1+q_2)^2}{4} = 0.625.

    Notice that

    9×1090.09=1×1011.\frac{9\times10^9}{0.09} = 1\times10^{11}.

    Thus,

    1×1011⋅(q1+q2)24=0.625⟹(q1+q2)2=0.625×41×1011=2.51×1011=2.5×10−11.1\times10^{11}\cdot\frac{ (q_1+q_2)^2}{4} = 0.625 \quad \Longrightarrow \quad (q_1+q_2)^2 = \frac{0.625\times4}{1\times10^{11}} = \frac{2.5}{1\times10^{11}} = 2.5\times10^{-11}.

    Taking the square root,

    q1+q2=5.0×10−6 C.q_1+q_2 = 5.0 \times10^{-6}\, \text{C}.

    Let Q=q1+q2=5.0×10−6 CQ = q_1+q_2 = 5.0 \times10^{-6}\, \text{C}.

  3. Finding the Individual Charges
    With q1+q2=5.0×10−6 Cq_1 + q_2 = 5.0\times10^{-6}\, \text{C} and knowing they are unequal with opposite signs, let

    q1=5.0×10−6+aandq2=−a,(a>0).q_1 = 5.0\times10^{-6} + a \quad \text{and} \quad q_2 = -a,\quad (a>0).

    Now, from the product:

    ∣q1q2∣=(5.0×10−6+a)a=1.4×10−11.|q_1q_2| = (5.0\times10^{-6} + a)a = 1.4\times10^{-11}.

    This gives the quadratic:

    a2+(5.0×10−6)a−1.4×10−11=0.a^2 + (5.0\times10^{-6})a - 1.4\times10^{-11} = 0.

    The discriminant DD is:

    D=(5.0×10−6)2+4(1.4×10−11)=25×10−12+5.6×10−11=8.1×10−11.D = (5.0\times10^{-6})^2 + 4(1.4\times10^{-11}) = 25\times10^{-12} + 5.6\times10^{-11} = 8.1\times10^{-11}.

    Hence,

    a=−5.0×10−6±8.1×10−112.a = \frac{-5.0\times10^{-6} \pm \sqrt{8.1\times10^{-11}}}{2}.

    Since 8.1×10−11=9.0×10−6\sqrt{8.1\times10^{-11}} = 9.0\times10^{-6} and a>0a>0, take the positive root:

    a=−5.0×10−6+9.0×10−62=4.0×10−62=2.0×10−6 C.a = \frac{-5.0\times10^{-6} + 9.0\times10^{-6}}{2} = \frac{4.0\times10^{-6}}{2} = 2.0\times10^{-6}\, \text{C}.

    Therefore,

    q1=5.0×10−6+2.0×10−6=7.0×10−6 C,q_1 = 5.0\times10^{-6} + 2.0\times10^{-6} = 7.0\times10^{-6}\, \text{C}, q2=−2.0×10−6 C.q_2 = -2.0\times10^{-6}\, \text{C}.

Explanation (Minimal):

  1. Use Coulomb’s law to get ∣q1q2∣|q_1q_2| from the initial attraction.
  2. After contact, the common charge is (q1+q2)/2(q_1+q_2)/2; use Coulomb’s law again with repulsive force to find q1+q2q_1+q_2.
  3. Solve q1+q2=5.0×10−6 Cq_1 + q_2 = 5.0\times10^{-6}\, \text{C} and (q1+a)(a)=1.4×10−11 C2(q_1+ a)(a)=1.4\times10^{-11}\, \text{C}^2 by setting q1=5.0×10−6+aq_1 = 5.0\times10^{-6}+a and q2=−aq_2=-a.
  4. Obtain a=2.0×10−6 Ca=2.0\times10^{-6}\, \text{C} yielding q1=7.0×10−6 Cq_1=7.0\times10^{-6}\, \text{C} and q2=−2.0×10−6 Cq_2=-2.0\times10^{-6}\, \text{C}.

Answer:

The initial charges are:
- q1=+7.0×10−6 Cq_1 = +7.0 \times 10^{-6}\, \text{C}
- q2=−2.0×10−6 Cq_2 = -2.0 \times 10^{-6}\, \text{C}