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(A) The electrostatic force provides the necessary centripetal force for circular motion: $F_e = \frac{mv^2}{r}$.Substituting the charges $q_1 = q$ an

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$\frac{3q^2}{2\varepsilon_0 h}$

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(A) The electrostatic force provides the necessary centripetal force for circular motion: $F_e = \frac{mv^2}{r}$.Substituting the charges $q_1 = q$ and $q_2 = 3q$ (taking magnitude for force),we get: $\frac{1}{4\pi\varepsilon_0} \cdot \frac{q(3q)}{r^2} = \frac{mv^2}{r}$.Simplifying this,we have: $\frac{3q^2}{4\pi\varepsilon_0 r} = mv^2$,which can be written as: $\frac{3q^2}{4\pi\varepsilon_0} = (mvr)v$.According to Bohr's quantization condition,the angular momentum is $mvr = \frac{nh}{2\pi}$. For the ground state,$n=1$,so $mvr = \frac{h}{2\pi}$.Substituting this into the force equation: $\frac{3q^2}{4\pi\varepsilon_0} = \left(\frac{h}{2\pi}\right)v$.Solving for $v$: $v = \frac{3q^2}{4\pi\varepsilon_0} \cdot \frac{2\pi}{h} = \frac{3q^2}{2\varepsilon_0 h}$.

In a hypothetical system,a particle of mass $m$ and charge $-3q$ is moving around a very heavy particle having charge $q$. Assuming Bohr's model to be true for this system,the orbital velocity of mass $m$ when it is in its ground state is:

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