(B) In equilibrium,the forces acting on one of the spherical balls are the tension $T$ in the thread,the electrostatic force $F_e$,and the gravitation

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(B) In equilibrium,the forces acting on one of the spherical balls are the tension $T$ in the thread,the electrostatic force $F_e$,and the gravitational force $mg$.Let the length of the thread be $L = 1\,m$ and the angle with the vertical be $\theta = 30^o$.The distance $r$ between the two charges is $r = 2L \sin(30^o) = 2 \times 1 \times 0.5 = 1\,m$.The electrostatic force is $F_e = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{r^2} = 9 \times 10^9 \times \frac{(10 \times 10^{-6})^2}{1^2} = 9 \times 10^9 \times 10^{-10} = 0.9\,N$.For equilibrium in the horizontal direction: $T \sin(30^o) = F_e$.$T \times 0.5 = 0.9$.$T = 1.8\,N$.

Class 12 Physics Electric Charges and Fields Electrostatic Force and Coulombs Law

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Two small spherical balls,each carrying a charge $Q = 10\,\mu C$ ($10$ micro-coulomb),are suspended by two insulating threads of equal lengths $1\,m$ each,from a point fixed in the ceiling. It is found that in equilibrium,the threads are separated by an angle of $60^o$ between them,as shown in the figure. What is the tension in the threads in $N$? (Given: $\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9\,Nm^2/C^2$)

A
$18$
B
$1.8$
C
$0.18$
D
None of the above

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Electric Charges and Fields

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Electrostatic Force and Coulombs Law

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DifficultJEE MAIN 2024

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DifficultTS EAMCET 2008

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The ratio of electrostatic and gravitational forces acting between an electron and a proton separated by a distance $5 \times 10^{-11} \, m$ will be (Charge on electron $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$,mass of proton $= 1.6 \times 10^{-27} \, kg$,$G = 6.7 \times 10^{-11} \, N m^2/kg^2$).

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In a hydrogen atom,the distance between the electron and proton is $2.5 \times 10^{-11} \ m$. The electrical force of attraction between them will be

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