Answer
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Hint:To answer this question, we first need to understand that in equilibrium the sum of all external forces acting on the body (first condition of equilibrium) and the sum of all external torques from external forces (second condition of equilibrium) are both needed for equilibrium (second condition of equilibrium). In order to achieve equilibrium, all of these conditions must be met at the same time.
Complete step by step answer:
Electric force: An electric force is the repulsive or enticing interaction between any two charged bodies. Newton's laws of motion explain the impact and consequences of any force on a given body, just as they do with any other force. The electric force is one of the many forces that exert themselves on objects.
Gravitational force: The force exerted by the Earth on you is equal to the force exerted by the Earth on you. The gravitational force equals your weight when you're at rest on or near the Earth's surface.
As given in the problem the charges are in equilibrium , therefore as per above definition of equilibrium net force is equal to zero.Hence,
${F_e} = {F_g}$
Where ${F_e}$ is the electric force between the charges and ${F_g}$ is the gravitational force between the charges. Therefore,
$\dfrac{1}{{4\pi { \in _0}}}\dfrac{{{Q_1}{Q_2}}}{{{R^2}}} = \dfrac{{G{M_1}{M_2}}}{{{R^2}}}$ (where $G$ is the gravitational constant )
As charges and mass of both the charges are equal
$\dfrac{{{Q^2}}}{{{M^2}}} = G \times 4\pi { \in _0}$
$\Rightarrow \dfrac{{{Q^2}}}{{{M^2}}} = \dfrac{{6.67 \times {{10}^{ - 11}}}}{{9 \times {{10}^9}}}$......(As G = $6.67 \times {10^{ - 11}}N{m^2}/K{g^2}$ )
We get,
$\therefore \dfrac{Q}{M} = 0.7411 \times {10^{ - 10}}$ C/Kg
So this is the final answer.
Note:The condition of an entity in which all of the forces acting on it are balanced is called equilibrium. The net force in such situations is 0 Newton. Trigonometric functions may be used to calculate the horizontal and vertical components of each force when the forces acting on an object are known.
Complete step by step answer:
Electric force: An electric force is the repulsive or enticing interaction between any two charged bodies. Newton's laws of motion explain the impact and consequences of any force on a given body, just as they do with any other force. The electric force is one of the many forces that exert themselves on objects.
Gravitational force: The force exerted by the Earth on you is equal to the force exerted by the Earth on you. The gravitational force equals your weight when you're at rest on or near the Earth's surface.
As given in the problem the charges are in equilibrium , therefore as per above definition of equilibrium net force is equal to zero.Hence,
${F_e} = {F_g}$
Where ${F_e}$ is the electric force between the charges and ${F_g}$ is the gravitational force between the charges. Therefore,
$\dfrac{1}{{4\pi { \in _0}}}\dfrac{{{Q_1}{Q_2}}}{{{R^2}}} = \dfrac{{G{M_1}{M_2}}}{{{R^2}}}$ (where $G$ is the gravitational constant )
As charges and mass of both the charges are equal
$\dfrac{{{Q^2}}}{{{M^2}}} = G \times 4\pi { \in _0}$
$\Rightarrow \dfrac{{{Q^2}}}{{{M^2}}} = \dfrac{{6.67 \times {{10}^{ - 11}}}}{{9 \times {{10}^9}}}$......(As G = $6.67 \times {10^{ - 11}}N{m^2}/K{g^2}$ )
We get,
$\therefore \dfrac{Q}{M} = 0.7411 \times {10^{ - 10}}$ C/Kg
So this is the final answer.
Note:The condition of an entity in which all of the forces acting on it are balanced is called equilibrium. The net force in such situations is 0 Newton. Trigonometric functions may be used to calculate the horizontal and vertical components of each force when the forces acting on an object are known.
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