Question

Suppose that the magnitude of the charge on the yellow sphere is determined to be \(2 q\). Calculate the charge \(q_{\text {red }}\) on the red sphere. Express your answer in terms of \(q, d_{1}, d_{2}\), and \(\theta\). qred =

Answer 1

Concepts and Reason

The concept used to solve the problem is coulomb’s law.

The nature of charge of yellow sphere can be calculated with the help of the direction of net force acting on the blue sphere.

The nature of charge on red sphere can thus be calculated by knowing the nature of charge of yellow and blue sphere.

Fundamentals

Like charges attract each other and unlike charges repel each other.

Coulomb’s Law: The magnitude of net electrical force between charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges.

The magnitude of the net electrical force between charges is calculated as follows,

$F = \left( {\frac{{k{q_1}{q_2}}}{{{r^2}}}} \right)$

Here, k is the force constant, q₁ is the charge of proton, q₂ is the charge of electron on the x-axis, and r is the distance between the charges.

A blue sphere at the origin with positive charge q and a red sphere fixed at the point $({d_1},0)$ with an unknown charge ${q_{{\rm{red}}}}$ , and a yellow sphere is fixed at point $({d_2}\cos \theta , - {d_2}\sin \theta )$ with unknown charge ${q_{{\rm{yellow}}}}$ .

The x component of the position vector of yellow sphere is positive, that is, ${d_2}\cos \theta$ and the y component of the position vector of yellow sphere is negative, that is, $- {d_2}\sin \theta$ . The net force on the blue sphere is observed to be ${F_{{\rm{vector}}}}$ placed at $(0,F)$ where F is greater than zero.

The net force on the charge blue sphere is along the negative y axis and the y component of positive vector of yellow sphere is negative, that is, $- {d_2}\sin \theta$ .

Hence, the yellow sphere must be attracting blue sphere, which means nature of charge on the yellow sphere is opposite to the nature of charge on the blue sphere. Therefore, the charge on the blue sphere is positive and the charge on the yellow sphere is negative.

As the component of net force on x - axis is equal to zero, the x component of force on blue sphere due to the yellow sphere must be equal and opposite to force on blue sphere due to red one.

‎In other words, as the yellow sphere attracts blue sphere and the red sphere must repel the blue sphere.

‎Hence, the nature of charge on ${q_{{\rm{red}}}}$ is opposite to nature of charge on ${q_{{\rm{yellow}}}}$ .
‎
‎Thus, the charge on yellow sphere is negative, hence charge on red is positive.

Since the charge on the yellow sphere is negative, and the net force on the blue sphere is along the negative y axis, the sign of the charge on the red sphere is positive.

‎As the x component of resultant force is equal to zero, therefore,

$\begin{array}{c}\\\sum {{F_{\rm{x}}}} = 0\\\\{F_{{\rm{x,yellow}}}}\cos \theta = {F_{{\rm{x,red}}}}\\\end{array}$

Substitute $\frac{{k\left( {2q} \right)\left( q \right)}}{{d_2^2}}$ for ${F_{{\rm{x,yellow}}}}$ and $\frac{{k{q_{{\rm{red}}}}\left( q \right)}}{{d_1^2}}$ for ${F_{{\rm{x,red}}}}$ in the above equation.

$\left( {\frac{{k\left( {2q} \right)\left( q \right)}}{{d_2^2}}} \right)\cos \theta = \frac{{k{q_{{\rm{red}}}}\left( q \right)}}{{d_1^2}}$

Rearrange the above equation.

${q_{{\rm{red}}}} = 2q\cos \theta {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}$

Ans:

The charge on the red sphere is equal to $2q\cos \theta {\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}$ .