Physics, asked by avinsmallo4312, 10 months ago

Two spherical bodies, each of mass 50 kg, are placed at a separation of 20 cm. Equal charges are placed on the bodies and it is found that the force of Coulomb repulsion equals the gravitational attraction in magnitude. Find the magnitude of the charge placed on either body.

Answers

Answered by bhuvna789456
2

The magnitude of the charge placed on either body is, q=4.3 \times 10^{-9}.

Explanation:

Step 1:

The given data in questions  :

                      Mass = m₁ = m₂ = 50 kg.

                     The distance ( r) = 20 cm        

Converting centimeter to meter  :

                                    \mathrm{r}=\frac{20}{100}=0.2 \mathrm{m}

                                   G=6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}

                                \frac{1}{4 \pi \varepsilon_{0}}=9.0 \times 10^{9} \mathrm{N}-\mathrm{m}^{2} / \mathrm{C}^{2}

Step 2:

Based on the question,

                                  F_{g}=F_{e}

Here,  

                  F_{g}=\frac{G m_{1} m_{2}}{r^{2}}   and    F_{e}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}          

                               \frac{G m_{1} m_{2}}{r^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}

                \frac{6.67 \times 10^{-11} \times 50 \times 50}{0.2^{2}}=9.0 \times 10^{9} \times \frac{q \times q}{r^{2}}

Where, \left(q_{1} q_{2}=q\right)

                 \frac{667 \times 10^{-11} \times 50 \times 50}{0.2^{2}}=9.0 \times 10^{9} \times \frac{q^{2}}{0.2^{2}}

                   \frac{667 \times 10^{-11} \times 2500}{0.04}=9.0 \times 10^{9} \times \frac{q^{2}}{0.04}

           6.67 \times 10^{-11} \times 2500=9.0 \times 10^{9} \times q^{2}

                   \frac{6.67 \times 10^{-11} \times 2500}{9.0 \times 10^{9}}=q^{2}

                                       q^{2}=\frac{6.67 \times 10^{-11} \times 2500}{9.0 \times 10^{9}}

                                       q^{2}=\frac{16675 \times 10^{-11}}{9.0 \times 10^{9}}

                                       q^{2}=1852.77 \times 10^{-20}

Step 3:        

Squaring on both sides, we get

                                         q=\sqrt{1852.77 \times 10^{-20}}

                                         q=\sqrt{1852.77 \times 10^{-20}}

                                         q=4.3 \times 10^{-9}

Thus, the magnitude of charge is, 4.3×10^-^9.

Answered by Anonymous
1

\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}

q = 4.3 \times  {10}^{ - 9}

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