Point charges of \[{\rm{5}}{\rm{.00 \mu C}}\] and \[{\rm{--3}}{\rm{.00 \mu C}}\] are placed \[{\rm{0}}{\rm{.250 m}}\] apart. (a) Where can a third charge be placed so that the net force on it is zero? (b) What if both charges are positive?

When a test charge is placed in a system of charges, each charge will exert electrostatic force on it. The vector sum of all the forces acting on the test charge is known as Net force or the resultant force.

(a)

The force representation on test charge \(q\) is,

Force on test charge

Here, \({F_1}\) is the repulsive force due to positively chargec, \({F_2}\) is the attractive force due to negative charge \({q_2}\), r is the separation between \({q_1}\) and \({q_2}\), and x is the separation between the test charge \(q\) and \({q_2}\).

The repulsive force due to positively charged \({q_1}\),

\({F_1} = \frac{{Kq{q_1}}}{{{{\left( {r + x} \right)}^2}}}\)

Here, K is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the test charge, \({q_1}\) is the magnitude of the positive charge \(\left( {{q_1} = 5.00{\rm{ }}\mu C} \right)\), ris the separation between \({q_1}\) and \({q_2}\) \(\left( {r = 0.250{\rm{ }}m} \right)\), and x is the separation between the test charge \(q\) and \({q_2}\).

The attractive force due to negatively charged \({q_2}\),

\({F_2} = \frac{{Kq{q_2}}}{{{x^2}}}\)

Here, K is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the test charge, \({q_2}\) is the magnitude of the negative charge \(\left( {{q_2} = 3.00{\rm{ \mu C}}} \right)\), and x is the separation between the test charge \(q\) and \({q_2}\).

Since, the net force on test charge is zero. Therefore,

\(\begin{aligned} {F_1} &= {F_2}\\\frac{{Kq{q_1}}}{{{{\left( {r + x} \right)}^2}}} &= \frac{{Kq{q_2}}}{{{x^2}}}\end{aligned}\)\(\)

The expression for the distance of the test charge \(q\) for negative charge \({q_2}\) is,

\(\begin{aligned} {\left( {\frac{{r + x}}{x}} \right)^2} &= \frac{{{q_1}}}{{{q_2}}}\\\frac{r}{x} + 1 &= \sqrt {\frac{{{q_1}}}{{{q_2}}}} \\\frac{r}{x} &= \sqrt {\frac{{{q_1}}}{{{q_2}}}} - 1\\x &= \frac{r}{{\sqrt {\frac{{{q_1}}}{{{q_2}}}} - 1}}\end{aligned}\)

Substituting all known values,

\[\begin{aligned} x &= \frac{{0.250{\rm{ m}}}}{{\sqrt {\frac{{5.00{\rm{ \mu C}}}}{{3.00{\rm{ \mu C}}}}} - 1}}\\ &= 0.859{\rm{ m}}\end{aligned}\]

Hence, the charge must be placed \[0.859{\rm{ m}}\] from the negative charge.

(b)

The force representation on test charge \(q\) is,

Force on test charge

Here, \({F_1}\) and \({F_2}\) are the repulsive force due to positive charges \({q_1}\) and \({q_2}\), \({\rm{r}}\) is the separation between \({q_1}\) and \({q_2}\), and \({\rm{x}}\) is the separation between the test charge \(q\) and \({q_1}\).

The repulsive force due to positively charged \({q_1}\),

\({F_1} = \frac{{Kq{q_1}}}{{{x^2}}}\)

Here, K is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the test charge, \({q_1}\) is the magnitude of the positive charge \(\left( {{q_1} = 5.00{\rm{ }}\mu C} \right)\), r is the separation between \({q_1}\) and \({q_2}\) \(\left( {r = 0.250{\rm{ }}m} \right)\), and x is the separation between the test charge \(q\) and \({q_1}\).

The repulsive force due to positively charges \({q_2}\),

\({F_2} = \frac{{Kq{q_2}}}{{{x^2}}}\)

Here, K is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), \(q\) is the test charge, \({q_2}\) is the magnitude of the positive charge \(\left( {{q_2} = 3.00{\rm{ \mu C}}} \right)\), r is the separation between \({q_1}\) and \({q_2}\) \(\left( {r = 0.250{\rm{ }}m} \right)\), and x is the separation between the test charge \(q\) and \({q_2}\).

Since, the net force on test charge is zero. Therefore,

\(\begin{aligned}{c}{F_1} = {F_2}\\\frac{{Kq{q_1}}}{{{x^2}}} = \frac{{Kq{q_2}}}{{{{\left( {r - x} \right)}^2}}}\end{aligned}\)\(\)

The expression for the distance of the test charge \(q\) for negative charge \({q_2}\) is,

\(\begin{aligned} {\left( {\frac{{r - x}}{x}} \right)^2} &= \frac{{{q_2}}}{{{q_1}}}\\\frac{r}{x} - 1 &= \sqrt {\frac{{{q_2}}}{{{q_1}}}} \\\frac{r}{x} &= \sqrt {\frac{{{q_2}}}{{{q_1}}}} + 1\\x &= \frac{r}{{\sqrt {\frac{{{q_2}}}{{{q_1}}}} + 1}}\end{aligned}\)

Substituting all known values,

\[\begin{aligned} x &= \frac{{0.250{\rm{ m}}}}{{\sqrt {\frac{{3.00{\rm{ \mu C}}}}{{5.00{\rm{ \mu C}}}}} + 1}}\\ &= 0.141{\rm{ m}}\end{aligned}\]

Hence, the charge must be placed \[0.141{\rm{ m}}\] from the positive charge.