Question

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywhere on the y-axis.

1) Find an expression for (Fnet)x, the x-component of the net force on q. (Give your answer in terms of Q, q, a, y and constant K.)

Answer 1

Concepts and reason

The concept required to solve the given problem is electrostatic force between two charges.

Calculate the electrostatic force due to the charges $Q{\rm{ and }} - Q$ on $q$ by using expression for electrostatic force. Then, take the vector sum of these forces to fine the net force on the charge q.

Fundamentals

According to Coulomb’s law, the force of attraction or repulsion between two charged particles is proportional to product magnitude of charge on each particle and inversely proportional to square of the distance between the charges.

The electrostatic force ${F_{AB}}$ between two charges is given as follows:

${F_{AB}} = \frac{{k{q_A}{q_B}}}{{{r_{AB}}^2}}$

Here, $k$ is the coulomb’s constant, ${q_A}{\rm{ and }}{q_B}$ are the charges and ${r_{AB}}$ is the distance between two charges ${q_A}{\rm{ and }}{q_B}$ .

Draw the following figure from given data.

From the figure, distance ${r_1}$ between charges q and Q can be calculated by using Pythagoras theorem as,

$\begin{array}{c}\\r_1^2 = {a^2} + {y^2}\\\\{r_1} = \sqrt {{a^2} + {y^2}} \\\end{array}$

From the figure, distance ${r_2}$ between charges q and -Q can be calculated by using Pythagoras theorem as,

$\begin{array}{c}\\r_2^2 = {a^2} + {y^2}\\\\{r_2} = \sqrt {{a^2} + {y^2}} \\\end{array}$

From the figure, the vector form of electrostatic force ${F_1}$ on charge q due to charge Q is given as follows:

$\begin{array}{c}\\{F_1} = {F_{1x}} + {F_{1y}}\\\\ = \left( { - \frac{{kqQ}}{{r_1^2}}\cos \theta \,{\rm{\hat i}} + \frac{{kqQ}}{{r_1^2}}\sin \theta \,{\rm{\hat j}}} \right)\\\\ = \frac{{kqQ}}{{r_1^2}}\left( { - \cos \theta \,{\rm{\hat i}} + \sin \theta \,{\rm{\hat j}}} \right)\\\end{array}$

From the figure, the vector form of electrostatic force ${F_2}$ on charge q due to charge -Q is given as follows:

$\begin{array}{c}\\{F_2} = {F_{2x}} + {F_{2y}}\\\\ = \left( { - \frac{{kqQ}}{{r_2^2}}\cos \theta \,{\rm{\hat i}} - \frac{{kqQ}}{{r_2^2}}\sin \theta \,{\rm{\hat j}}} \right)\\\\ = \frac{{kqQ}}{{r_2^2}}\left( { - \cos \theta \,{\rm{\hat i}} - \sin \theta \,{\rm{\hat j}}} \right)\\\end{array}$

The net force F on q is equal to the sum of forces ${F_1}$ and ${F_2}.$

$F = {F_1} + {F_2}$

Substitute $\frac{{kqQ}}{{r_1^2}}\left( { - \cos \theta \,{\rm{\hat i}} + \sin \theta \,{\rm{\hat j}}} \right)$ for ${F_1},$ and $\frac{{kqQ}}{{r_2^2}}\left( { - \cos \theta \,{\rm{\hat i}} - \sin \theta \,{\rm{\hat j}}} \right)$ for ${F_2}$ in the above equation.

$F = \frac{{kqQ}}{{r_1^2}}\left( { - \cos \theta \,{\rm{\hat i}} + \sin \theta \,{\rm{\hat j}}} \right) + \frac{{kqQ}}{{r_2^2}}\left( { - \cos \theta \,{\rm{\hat i}} - \sin \theta \,{\rm{\hat j}}} \right)$

Substitute $\sqrt {{a^2} + {y^2}}$ for ${r_1}$ and ${r_2}$ in the above equation.

$\begin{array}{c}\\F = \frac{{kqQ}}{{{{\left( {\sqrt {{a^2} + {y^2}} } \right)}^2}}}\left( { - \cos \theta \,{\rm{\hat i}} + \sin \theta \,{\rm{\hat j}}} \right) + \frac{{kqQ}}{{{{\left( {\sqrt {\left( {{a^2} + {y^2}} \right)} } \right)}^2}}}\left( { - \cos \theta \,{\rm{\hat i}} - \sin \theta \,{\rm{\hat j}}} \right)\\\\ = \frac{{kqQ}}{{\left( {{a^2} + {y^2}} \right)}}\left( {\left( { - \cos \theta - \cos \theta } \right)\,{\rm{\hat i}} + \left( {\sin \theta - \sin \theta } \right)\,{\rm{\hat j}}} \right)\\\\ = \frac{{ - 2kqQ}}{{\left( {{a^2} + {y^2}} \right)}}\cos \theta \,{\rm{\hat i}}\\\end{array}$

From the figure, the values of $\cos \theta$ is,

$\begin{array}{c}\\\cos \theta = \frac{a}{{{r_1}}}\\\\ = \frac{a}{{\sqrt {{a^2} + {y^2}} }}\\\end{array}$

Substitute $\frac{a}{{\sqrt {{a^2} + {y^2}} }}$ for $\cos \theta$ in the above equation and solve for F.

$\begin{array}{c}\\F = - \frac{{2kqQ}}{{\left( {{a^2} + {y^2}} \right)}}\left( {\frac{a}{{\sqrt {{a^2} + {y^2}} }}} \right)\,{\rm{\hat i}}\\\\{\rm{ = }} - \frac{{2kqQa}}{{{{\left( {{a^2} + {y^2}} \right)}^{3/2}}}}{\rm{\hat i}}\\\end{array}$

Ans:

The x-component of force on the charge q is $- \frac{{2kqQa}}{{{{\left( {{a^2} + {y^2}} \right)}^{3/2}}}}.$