Question

A small sphere of charge q₁ = 0.864 µC hangs from the end of a spring as in Figure a. When another small sphere of charge q₂ = -0.66 µC is held beneath the first sphere as in Figure b, the spring stretches by d = 3.27 cm from its original length and reaches a new equilibrium position with a separation between the charges of r = 4.80 cm. What is the force constant of the spring?
N/m

Answer 1

Concepts and reason

The concepts required to solve this problem are the Hooke’s law and the electric force.

Initially, write the force equation for the system in the figure$\left( {\rm{a}} \right)$ using the Hooke’s law. Then, write the force equation for the system in the figure$\left( {\rm{b}} \right)$ using the Hooke’s law and the electric force for the new equilibrium position. Simplify the equation and calculate the spring constant.

Fundamentals

The expression of the force ${F_{\rm{s}}}$ using the Hooke’s law is,

${F_{\rm{s}}} = - kx$

Here, ${F_{\rm{s}}}$ is the force, $k$ is the spring constant, and x is the displacement of the spring.

The expression of the coulomb’s force is,

${F_{\rm{e}}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$

Here, ${F_{\rm{e}}}$ is the Coulomb’s force, ${\varepsilon _0}$ is the permittivity of free space, r is the distance, ${q_1}$ and ${q_2}$ are the charges.

The force ${F_{\rm{g}}}$ of an object due to gravity is,

${F_{\rm{g}}} = mg$

Here, m is the mass of the object and g is the acceleration due to gravity.

Refer figure$\left( {\rm{a}} \right)$ given in the question.

The expression for the force is given by,

${F_{\rm{s}}} = k{x_1}$

Here, ${x_1}$ is the displacement.

The equation of the force in this equilibrium position is,

${F_{\rm{s}}} = {F_{\rm{g}}}$

Substitute $k{x_1}$ for ${F_{\rm{s}}}$ and mg for ${F_{\rm{g}}}$.

$k{x_1} = mg$

Refer figure$\left( {\rm{b}} \right)$ given in the question.

The equation of the force in this new equilibrium position due to the presence of the other charge ${q_2}$ is,

$k\left( {d + {x_1}} \right) = mg + {F_{\rm{e}}}$

Here, $\left( {d + {x_1}} \right)$ is the stretched length of the spring.

Simplify the above equation as follows:

$\begin{array}{c}\\k\left( {d + {x_1}} \right) = mg + {F_{\rm{e}}}\\\\kd + k{x_1} = mg + {F_{\rm{e}}}\\\end{array}$

Substitute $k{x_1}$ for $mg$ in above equation as follows:

$\begin{array}{c}\\kd + k{x_1} = k{x_1} + {F_{\rm{e}}}\\\\kd = {F_{\rm{e}}}\\\end{array}$

The spring constant using the above equation is,

$k = \frac{{{F_{\rm{e}}}}}{d}$

The expression of the coulomb’s force is,

${F_{\rm{e}}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$

Since both the charges are of opposite sign so the force is attractive in nature.

Substitute $0.864{\rm{ }}\mu {\rm{C}}$ for ${q_1}$, $- 0.66{\rm{ }}\mu {\rm{C}}$ for ${q_2}$, $8.85 \times {10^{ - 12}}{\rm{ F/m}}$ for ${\varepsilon _0}$, and 4.80 cm for r.

$\begin{array}{c}\\{F_{\rm{e}}} = \frac{{\left( {0.864{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)} \right)\left( {0.66{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)} \right)}}{{4\pi \left( {8.85 \times {{10}^{ - 12}}{\rm{ F/m}}} \right){{\left( {4.80{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^2}}}\\\\{\rm{ = 2}}{\rm{.2275 N}}\\\end{array}$

The spring constant is,

$k = \frac{{{F_{\rm{e}}}}}{d}$

Substitute 2.2275 N for ${F_{\rm{e}}}$ and 3.27 cm for d.

$\begin{array}{c}\\k = \frac{{2.2275{\rm{ N}}}}{{3.27{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)}}\\\\ = 68.1{\rm{ N/m}}\\\end{array}$

Ans:

The force constant of the spring is $68.1{\rm{ N/m}}$.