Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0765 N. Then, a thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed the spheres still repel but with a force of F2 = 0.100 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1

Solve q1 and q2

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Answer 1

Concepts and reason

The concepts used to solve this problem are electrostatic force, flow of charge.

Use the concept of electrostatic force to calculate the relation between unknown charges at the initial condition.

Use the concept of flow of charge and electrostatic force to calculate the relation between unknown charges after redistribution of charge.

Finally, solve both the relation between charge during initial condition and after redistributing of charge find the value of unknown charges.

Fundamentals

Write the expression for the electrostatic force.

$F = k\frac{{{q_1}{q_2}}}{{{r^2}}}$

Here, ${q_1}$ and ${q_2}$ are two charges, $r$ is the distance between the charge and $k$ is a constant whose value is $8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}$ .

Write the concept of charge distribution.

Charge always flow from higher to lower amount of charge. when two charges having higher and lower value are connected with conducting wire the charge will flow higher to lower amount of charge till both the charge are equal.

Write the expression for the electrostatic force.

$F = k\frac{{{q_1}{q_2}}}{{{r^2}}}$

Here, ${q_1}$ and ${q_2}$ are two charges, $r$ is the distance between the charge and $k$ is a constant whose value is $8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}$ .

Substitute for $k$ , $44.0 {\rm{cm}}$ for $r$ and $0.0765 {\rm{N}}$ for $F$ .

$\begin{array}{l}\\0.0765 {\rm{N}} = \left( {8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{q_1}{q_2}}}{{{{\left( {44.0 {\rm{cm}}} \right)}^2}}}\\\\0.0765 {\rm{N}} = \left( {8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{q_1}{q_2}}}{{{{\left\{ {\left( {44.0 {\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}} {\rm{m}}}}{{1 {\rm{cm}}}}} \right)} \right\}}^2}}}\\\\0.0765 {\rm{N}} = \left( {8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{q_1}{q_2}}}{{{{\left( {44.0 \times {{10}^{ - 2}} {\rm{m}}} \right)}^2}}}\\\end{array}$

Solve further as,

$\begin{array}{l}\\{q_1}{q_2} = 1.65 \times {10^{ - 12}} {{\rm{C}}^2}\\\\{q_2} = \frac{{1.65 \times {{10}^{ - 12}} {{\rm{C}}^2}}}{{{q_1}}}\\\end{array}$ …… (1)

Find the charge on the sphere after redistributing the charge on each sphere.

Consider the charge on each sphere after redistributing the charge is $Q$ . Because, when two charges having higher and lower value are connected with conducting wire the charge will flow higher to lower value of charge till both the charge are equal.

Thus, ${q_1} = {q_2} = Q$

So,

$Q = \frac{{{q_1} + {q_2}}}{2}$

Now calculate the force between the sphere after redistribution of charge.

${q_1} = {q_2} = Q$

Write the expression for the electrostatic force.

$\begin{array}{c}\\F = k\frac{{QQ}}{{{r^2}}}\\\\ = k\frac{{{Q^2}}}{{{r^2}}}\\\end{array}$

Here, ${q_1}$ and ${q_2}$ are two charges, $r$ is the distance between the charge and $k$ is a constant whose value is $8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}$ .

Substitute $8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}$ for $k$ , for $r$ , $0.100 {\rm{N}}$ for $F$ , $\frac{{{q_1} + {q_2}}}{2}$ for $Q$ .

$\begin{array}{l}\\0.100 {\rm{N}} = \left( {8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{{\left( {\frac{{{q_1} + {q_2}}}{2}} \right)}^2}}}{{{{\left( {44.0 {\rm{cm}}} \right)}^2}}}\\\\0.100 {\rm{N}} = \left( {8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{{\left( {{q_1} + {q_2}} \right)}^2}}}{{4{{\left\{ {\left( {44.0 {\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}} {\rm{m}}}}{{1 {\rm{cm}}}}} \right)} \right\}}^2}}}\\\\0.400 {\rm{N}} = \left( {8.99 \times 1{0^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{{\left( {{q_1} + {q_2}} \right)}^2}}}{{{{\left( {44.0 \times {{10}^{ - 2}} {\rm{m}}} \right)}^2}}}\\\end{array}$

Solve further,

$\begin{array}{l}\\{\left( {{q_1} + {q_2}} \right)^2} = 8.6 \times {10^{ - 12}} {{\rm{C}}^2}\\\\\left( {{q_1} + {q_2}} \right) = \sqrt {8.6 \times {{10}^{ - 12}} {{\rm{C}}^2}} \\\end{array}$

${q_1} + {q_2} = 2.9 \times {10^{ - 6}} {\rm{C}}$ …… (2)

Calculate the value of charge.

Substitute $\frac{{1.65 \times {{10}^{ - 12}} {{\rm{C}}^2}}}{{{q_1}}}$ for ${q_2}$ from equation (1) in the equation (2)

$\begin{array}{l}\\{q_1} + \frac{{1.65 \times {{10}^{ - 12}} {{\rm{C}}^2}}}{{{q_1}}} = 2.9 \times {10^{ - 6}} {\rm{C}}\\\\{q_1}^2 + 1.65 \times {10^{ - 12}} {{\rm{C}}^2} = \left( {2.9 \times {{10}^{ - 6}} {\rm{C}}} \right){q_1}\\\end{array}$

Solve further as,

${q_1}^2 - \left( {2.9 \times {{10}^{ - 6}} {\rm{C}}} \right){q_1} + 1.65 \times {10^{ - 12}} {{\rm{C}}^2} = 0$

By solving the quadratic equation.

The roots of the equation are,

${q_1} = 2.12 \times {10^{ - 6}} {\rm{C}}$

And, ${q_1} = {\rm{7}}{\rm{.77}} \times {\rm{1}}{{\rm{0}}^{ - 7}} {\rm{C}}$ .

Now calculate ${q_2}$ .

If ${q_1} = 2.12 \times {10^{ - 6}} {\rm{C}}$ then from equation (1).

$\begin{array}{c}\\{q_2} = \frac{{1.65 \times {{10}^{ - 12}} {{\rm{C}}^2}}}{{2.12 \times {{10}^{ - 6}} {\rm{C}}}}\\\\ = 7.78 \times {10^{ - 7}} {\rm{C}}\\\end{array}$

And if ${q_1} = {\rm{7}}{\rm{.77}} \times {\rm{1}}{{\rm{0}}^{ - 7}} {\rm{C}}$ then from equation (1).

$\begin{array}{c}\\{q_2} = \frac{{1.65 \times {{10}^{ - 12}} {{\rm{C}}^2}}}{{{\rm{7}}{\rm{.77}} \times {\rm{1}}{{\rm{0}}^{ - 7}} {\rm{C}}}}\\\\ = 2.12 \times {10^{ - 6}} {\rm{C}}\\\end{array}$

Consider condition provided in the question, that is ${q_2} > {q_1}$ . Thus, the initial charge on each sphere ${q_1}$ and ${q_2}$ are $7.78 \times {10^{ - 7}} {\rm{C}}$ and $2.12 \times {10^{ - 6}} {\rm{C}}$ respectively.

Take reference from the previous step and understand that for the condition ${q_2} > {q_1}$ , the initial charge on second sphere ${q_2}$ is $2.12 \times {10^{ - 6}} {\rm{C}}$ respectively.

Ans:

The initial charge on first sphere ${q_1}$ is $7.78 \times {10^{ - 7}} {\rm{C}}$ .

The initial charge on second sphere ${q_2}$ is $2.12 \times {10^{ - 6}} {\rm{C}}$ .