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Two spheres have charges q1 = q2 = 6.5 imes 10⁻⁷ C and are separated by r = 50 cm. Calculate the electrostatic force. If the charges are doubled and the distance is halved, what is the new force?

Physics

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Answer

F=1.52×10−2 NF = \mathbf{1.52 \times 10^{-2} \text{ N}}

F′=0.243 NF' = \mathbf{0.243 \text{ N}}

Step-by-step solution

5 steps

  1. Identify the given values for part (a).

    The charges on the spheres are q1=q2=6.5×10−7 Cq_1 = q_2 = 6.5 \times 10^{-7} \text{ C}.
    The distance between the spheres is r=50 cm=0.50 mr = 50 \text{ cm} = 0.50 \text{ m}.
    The electrostatic constant is k=14πϵ0=9×109 N m2/C2k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2.

  2. Calculate the electrostatic force for part (a) using Coulomb's Law.

    The formula for Coulomb's Law is F=kq1q2r2F = k \frac{q_1 q_2}{r^2}.

    F=(9×109 N m2/C2)(6.5×10−7 C)×(6.5×10−7 C)(0.50 m)2F = (9 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{(6.5 \times 10^{-7} \text{ C}) \times (6.5 \times 10^{-7} \text{ C})}{(0.50 \text{ m})^2} F=(9×109)42.25×10−140.25 NF = (9 \times 10^9) \frac{42.25 \times 10^{-14}}{0.25} \text{ N} F=(9×109)×(169×10−14) NF = (9 \times 10^9) \times (169 \times 10^{-14}) \text{ N} F=1521×10−5 NF = 1521 \times 10^{-5} \text{ N} F=1.521×10−2 NF = 1.521 \times 10^{-2} \text{ N}

    Rounding to two decimal places for the coefficient:

    F=1.52×10−2 NF = \boxed{\mathbf{1.52 \times 10^{-2} \text{ N}}}

  3. Identify the new conditions for part (b).

    The new charges are q1′=2q1q_1' = 2q_1 and q2′=2q2q_2' = 2q_2.
    The new distance is r′=r2r' = \frac{r}{2}.

  4. Calculate the new electrostatic force for part (b).

    The new force F′F' is given by:

    F′=kq1′q2′(r′)2F' = k \frac{q_1' q_2'}{(r')^2}

    Substitute the new values:

    F′=k(2q1)(2q2)(r2)2F' = k \frac{(2q_1)(2q_2)}{\left(\frac{r}{2}\right)^2} F′=k4q1q2r24F' = k \frac{4q_1 q_2}{\frac{r^2}{4}} F′=16(kq1q2r2)F' = 16 \left( k \frac{q_1 q_2}{r^2} \right)

    Since F=kq1q2r2F = k \frac{q_1 q_2}{r^2}, we have F′=16FF' = 16F.

  5. Substitute the value of FF from part (a) into the expression for F′F'.

    F′=16×(1.52×10−2 N)F' = 16 \times (1.52 \times 10^{-2} \text{ N}) F′=24.32×10−2 NF' = 24.32 \times 10^{-2} \text{ N} F′=0.2432 NF' = 0.2432 \text{ N}

    Rounding to three decimal places:

    F′=0.243 NF' = \boxed{\mathbf{0.243 \text{ N}}}
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