Question

Two point charges are separated by 25.0cm(see the figure(Figure 1) ).

Find the magnitude of the net electric field these charges produce at point A.

Find the magnitude of the net electric field these charges produce at point B.

What would be the magnitude of the electric force this combination of charges would produce on a proton at A?

Answer 1

Concepts and reason

The required concepts to solve the problem are electric field and electric force.

Initially, find the net electric field at point A by combining the fields produced by both charges at point A. Later, the net field at point B is found by combining the fields by both charges at point B. Finally, find the electric force produced by these charges on a proton at point A using the relation between charge, electric field and electric force.

Fundamentals

Electric field is a region around a charged particle where a charged particle experiences a force. The electric field is directly proportional to the charge and inversely proportional to the square of the distance between the point where the field is to be determined and the charge.

$E = \frac{{kQ}}{{{r^2}}}$

Here, $Q$is the charge and $r$is the distance between the charge and the point.

The force experienced by a charged particle in an electric field is the electric force. Electric force is the product of the charge and the electric field.

$F = qE$

Here, $q$is the charge and$E$is the electric field.

Electric field at point A is the difference of electric field due to the two charges. If one charge is${q_1}$and the other charge is${q_2}$, then the net electric field is,

$E = \frac{{k{q_1}}}{{r_1^2}} - \frac{{k{q_2}}}{{r_2^2}}$

Where,${r_1}$is the distance between the first charge and the point A and${r_2}$is the distance between the second charge and the point A.

$\begin{array}{c}\\{r_1} = \left( {\left( {25\,{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)} \right) - \left( {\left( {10\,{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)} \right)\\\\ = 0.15\,{\rm{m}}\\\end{array}$

And,

$\begin{array}{c}\\{r_2} = \left( {{\rm{10}}\,{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)\\\\ = 0.10\,{\rm{m}}\\\end{array}$

Substitute$9 \times {10^9}\,{\rm{N}} \cdot {{\rm{m}}^2}{{\rm{C}}^{ - 2}}$ for $k$,$- 6.25\,{\rm{nC}}$ for ${q_1}$,$- 12.5\,{\rm{nC}}$ for ${q_2}$, $0.15\,{\rm{m}}$ for ${r_1}$ and$0.10\,{\rm{m}}$ for ${r_2}$.

$\begin{array}{c}\\E = k\left( {\frac{{{q_1}}}{{r_1^2}} - \frac{{{q_2}}}{{r_2^2}}} \right)\\\\ = \left( {9 \times {{10}^9}\,{\rm{N}} \cdot {{\rm{m}}^2}{{\rm{C}}^{ - 2}}} \right)\left( {\left( {\frac{{\left( { - 6.25\,{\rm{nC}}} \right)\left( {\frac{{{{10}^{ - 9}}\,{\rm{C}}}}{{1\,{\rm{nC}}}}} \right)}}{{{{\left( {0.15\,{\rm{m}}} \right)}^2}}}} \right) - \left( {\frac{{\left( { - 12.5\,{\rm{nC}}} \right)\left( {\frac{{{{10}^{ - 9}}\,{\rm{C}}}}{{1\,{\rm{nC}}}}} \right)}}{{{{\left( {0.10\,{\rm{m}}} \right)}^2}}}} \right)} \right)\\\\ = 8750\,{\rm{N/C}}\,\\\end{array}$

Electric field at point B is the sum of electric field due to the two charges. If one charge is${q_1}$and the other charge is${q_2}$, then the net electric field is,

$E = \frac{{k{q_1}}}{{r_1^2}} + \frac{{k{q_2}}}{{r_2^2}}$

Where,${r_1}$is the distance between the first charge and the point B and${r_2}$is the distance between the second charge and the point B.

$\begin{array}{c}\\{r_2} = \left( {\left( {25\,{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)} \right) + \left( {\left( {10\,{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)} \right)\\\\ = 0.35\,{\rm{m}}\\\end{array}$

And,

$\begin{array}{c}\\{r_1} = \left( {{\rm{10}}\,{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\,{\rm{m}}}}{{1\,{\rm{cm}}}}} \right)\\\\ = 0.10\,{\rm{m}}\\\end{array}$

Substitute$9 \times {10^9}\,{\rm{N}} \cdot {{\rm{m}}^2}{{\rm{C}}^{ - 2}}$ for $k$,$- 6.25\,{\rm{nC}}$ for ${q_1}$,$- 12.5\,{\rm{nC}}$for${q_2}$, $0.10\,{\rm{m}}$for ${r_1}$ and$0.35\,{\rm{m}}$for ${r_2}$.

$\begin{array}{c}\\E = k\left( {\frac{{{q_1}}}{{r_1^2}} + \frac{{{q_2}}}{{r_2^2}}} \right)\\\\ = \left( {9 \times {{10}^9}\,{\rm{N}} \cdot {{\rm{m}}^2}{{\rm{C}}^{ - 2}}} \right)\left( {\left( {\frac{{\left( { - 6.25\,{\rm{nC}}} \right)\left( {\frac{{{{10}^{ - 9}}\,{\rm{C}}}}{{1\,{\rm{nC}}}}} \right)}}{{{{\left( {0.10\,{\rm{m}}} \right)}^2}}}} \right) + \left( {\frac{{\left( { - 12.5\,{\rm{nC}}} \right)\left( {\frac{{{{10}^{ - 9}}\,{\rm{C}}}}{{1\,{\rm{nC}}}}} \right)}}{{{{\left( {0.35\,{\rm{m}}} \right)}^2}}}} \right)} \right)\\\\ = - 6543.36\,{\rm{N/C}}\\\end{array}$

If the electric field at a point is known, then the electric force on a charge on that point is,

$F = qE$

Electric filed at point A is,

$E = 8750\,{\rm{N/C}}$

The charge of the proton is,

$q = 1.6 \times {10^{ - 19}}\,{\rm{C}}$

Substitute$8750\,{\rm{N/C}}$for$E$and$1.6 \times {10^{ - 19}}\,{\rm{C}}$for$q$to find$F$.

$\begin{array}{c}\\F = \left( {1.6 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( {8750\,{\rm{N/C}}} \right)\\\\ = 1.4 \times {10^{ - 15}}\,{\rm{N}}\\\end{array}$

Ans:

The magnitude of the net electric field these charges produce at point A is$8750\,{\rm{N/C}}$.

The magnitude of the net electric field these charges produce at point B is$6543.36\,{\rm{N/C}}$.

The electric force on a proton at point A is $1.4 \times {10^{ - 15}}\,{\rm{N}}$.