(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

Two positively charged spheres of masses $m_1$ and $m_2$ are suspended from a common point at the ceiling by identical insulating massless strings of length $l$. Charges on the two spheres are $q_1$ and $q_2$,respectively. At equilibrium,both strings make the same angle $\theta$ with the vertical. Then

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The force between two identical spheres charged with the same charge $q$ is $F$. If $50\%$ of the charge from one sphere is transferred to the second sphere,then the new force will be:

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Two identical pendulums $A$ and $B$ are suspended from the same point. The bobs are given positive charges,with $A$ having more charge than $B$. They diverge and reach equilibrium,with $A$ and $B$ making angles $\theta_1$ and $\theta_2$ with the vertical,respectively. Then:

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Four charges equal to $-Q$ are placed at the four corners of a square of side length $a$,and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is:

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Two point charges $+Q$ and $+q$ repel each other with a force of $100 \,N$. Keeping the distance between them unchanged, if $Q$ is increased by $10 \%$ and $q$ is decreased by $10 \%$, the force of repulsion between them will:

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The force between two identical spheres charged with the same charge $q$ is $F$. If $50\%$ of the charge from one sphere is transferred to the second sphere,then the new force will be:

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