Question

(a) Find the electric field at\(x = 5.00{\rm{ cm}}\)in Figure 18.52 (a), given that\(q = 1.00{\rm{ }}\mu C\). (b) At what position between\(3.00\)and\(8.00{\rm{ cm}}\)is the total electric field the same as that for\( - 2q\)alone? (c) Can the electric field be zero anywhere between\(0.00\)and\(8.00{\rm{ cm}}\)? (d) At very large positive or negative values of\(x\), the electric field approaches zero in both (a) and (b). In which does it most rapidly approach zero and why? (e) At what position to the right of\(11.0{\rm{ cm}}\)is the total electric field zero, other than at infinity? (Hint: A graphing calculator can yield considerable insight in this problem.)

Figure 18.52 (a) Point charges located at\[{\bf{3}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{11}}.{\bf{0}}{\rm{ }}{\bf{cm}}\]along the x-axis. (b) Point charges located at\[{\bf{1}}.{\bf{00}},{\rm{ }}{\bf{5}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{14}}.{\bf{0}}{\rm{ }}{\bf{cm}}\]along the x-axis

Short answer

(a) The electric field at \(x = 5.00{\rm{ cm}}\) is \(4.0 \times {10^7}{\rm{ N}}/{\rm{C}}\). (b) The electric field at \(x = 7.00{\rm{ cm}}\) will be same for the \( - 2q\) alone. (c) No, the electric field cannot be zero anywhere between \(0.00\) and \(8.00{\rm{ cm}}\). (d) In Figure 18.52 (a) the electric field rapidly approach zero. (e) The total electric field is zero at \(30.607{\rm{ cm}}\).

Step-by-step solution

Electric field

The region around a charge in which the effect of the charge can be observed is known as electric field.The expression for the electric field is,

\(E = \frac{{Kq}}{{{r^2}}}\)

Here, \(K\) is the electrostatic force constant, \(q\) is the charge and \(r\) is the distance of the point of observation from the charge.

(a) Electric field at x=5.00 cm

The electric field for Figure 18.52 at \(x = {\rm{5}}{\rm{.00 cm}}\) is represented as,

Electric field at\(x = 5.00{\rm{ cm}}\)

The electric field at\(x = 5.00{\rm{ cm}}\)is,

\(\begin{array}{c}E = {E_3} + {E_8} - {E_{11}}\\ = \frac{{K\left( q \right)}}{{{{\left| {{r_5} - {r_3}} \right|}^2}}} + \frac{{K\left( {2q} \right)}}{{{{\left| {{r_5} - {r_8}} \right|}^2}}} - \frac{{K\left( q \right)}}{{{{\left| {{r_5} - {r_{11}}} \right|}^2}}}\\ = K\left[ {\frac{{\left( q \right)}}{{{{\left| {{r_5} - {r_3}} \right|}^2}}} + \frac{{\left( {2q} \right)}}{{{{\left| {{r_5} - {r_8}} \right|}^2}}} - \frac{{\left( q \right)}}{{{{\left| {{r_5} - {r_{11}}} \right|}^2}}}} \right]\end{array}\)

Here, \(K\) is the electrostatic force constant, \(q\) is the charge, \(\left| {{r_5} - {r_3}} \right|\) is the distance between \(x = 5.00{\rm{ cm}}\) and the charge at \[x = 3.00{\rm{ cm}}\], \(\left| {{r_5} - {r_8}} \right|\) is the distance between \(x = 5.00{\rm{ cm}}\) and the charge at \(x = 8.00{\rm{ cm}}\), and \(\left| {{r_5} - {r_{11}}} \right|\) is the distance between \(x = 5.00{\rm{ cm}}\) and the charge at \(x = 11.00{\rm{ cm}}\).

Substitute \({\rm{9 \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}\) for \(K\), \(1.00{\rm{ }}\mu {\rm{C}}\) for \(q\)a, \(3.00{\rm{ cm}}\) for \({r_3}\), \(5.00{\rm{ cm}}\) for \({r_5}\), \(8.00{\rm{ cm}}\) for \({r_8}\), and \(11.00{\rm{ cm}}\) for \({r_{11}}\).

\(\begin{array}{c}E = \left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left[ \begin{array}{l}\frac{{\left( {{\rm{1}}{\rm{.00 \mu C}}} \right)}}{{{{\left| {\left( {{\rm{5}}{\rm{.00 cm}}} \right) - \left( {{\rm{3}}{\rm{.00 cm}}} \right)} \right|}^{\rm{2}}}}} + \frac{{{\rm{2 \times }}\left( {{\rm{1}}{\rm{.00 \mu C}}} \right)}}{{{{\left| {\left( {{\rm{5}}{\rm{.00 cm}}} \right) - \left( {{\rm{8}}{\rm{.00 cm}}} \right)} \right|}^{\rm{2}}}}}\\ - \frac{{\left( {{\rm{1}}{\rm{.00 \mu C}}} \right)}}{{{{\left| {\left( {{\rm{5}}{\rm{.00 cm}}} \right) - \left( {{\rm{11}}{\rm{.00 cm}}} \right)} \right|}^{\rm{2}}}}}\end{array} \right]\\ = \left( {{\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}} \right) \times \left[ \begin{array}{l}\frac{{\left( {{\rm{1}}{\rm{.00 \mu C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right)}}{{{{\left| {\left( {{\rm{2}}{\rm{.00 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right|}^{\rm{2}}}}} + \frac{{{\rm{2}} \times \left( {{\rm{1}}{\rm{.00 \mu C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right)}}{{{{\left| {\left( {{\rm{3}}{\rm{.00 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right|}^{\rm{2}}}}}\\ - \frac{{\left( {{\rm{1}}{\rm{.00 \mu C}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ C}}}}{{{\rm{1 \mu C}}}}} \right)}}{{{{\left| {\left( {{\rm{6}}{\rm{.00 cm}}} \right) \times \left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}{\rm{ m}}}}{{{\rm{1 cm}}}}} \right)} \right|}^{\rm{2}}}}}\end{array} \right]\\ = {\rm{4}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{7}}}{\rm{ N}}/{\rm{C}}\end{array}\)

Hence, the electric field at \(x = 5.00{\rm{ cm}}\) is \(4.0 \times {10^7}{\rm{ N}}/{\rm{C}}\).

(b) Point where the total electric field is same as that for -2q alone

Let \(x\) be the point between \(3.00\) and \(8.00{\rm{ cm}}\) where the total electric field be the same for \( - 2q\) alone i.e, \(E = {E_8}\).

The electric field at \(x\) is represented as,

Electric field at \(x\)

The electric field at \(x\) is,

\[\begin{array}{c}E = {E_3} + {E_8} - {E_{11}}\\{E_8} = \frac{{K\left( q \right)}}{{{{\left( {x - {r_3}} \right)}^2}}} + \frac{{K\left( {2q} \right)}}{{{{\left( {{r_8} - x} \right)}^2}}} - \frac{{K\left( q \right)}}{{{{\left( {{r_{11}} - x} \right)}^2}}}\\\frac{{K\left( {2q} \right)}}{{{{\left( {{r_8} - x} \right)}^2}}} = K\left[ {\frac{{\left( q \right)}}{{{{\left( {x - {r_3}} \right)}^2}}} + \frac{{\left( {2q} \right)}}{{{{\left( {{r_8} - x} \right)}^2}}} - \frac{{\left( q \right)}}{{{{\left( {{r_{11}} - x} \right)}^2}}}} \right]\\\frac{{K\left( {2q} \right)}}{{{{\left( {{r_8} - x} \right)}^2}}} = \frac{{\left( q \right)}}{{{{\left( {x - {r_3}} \right)}^2}}} + \frac{{\left( {2q} \right)}}{{{{\left( {{r_8} - x} \right)}^2}}} - \frac{{\left( q \right)}}{{{{\left( {{r_{11}} - x} \right)}^2}}}\end{array}\]

Here, \(K\) is the electrostatic force constant, \(q\) is the charge, \(\left( {x - {r_3}} \right)\) is the distance between \(x\) and the charge at \(x = {\rm{3}}{\rm{.00 cm}}\), \(\left( {{r_8} - x} \right)\) is the distance between \(x\) and the charge at \(x = 8.00{\rm{ cm}}\), and \(\left( {{r_{11}} - x} \right)\) is the distance between \(x\) and the charge at \(x = {\rm{11}}{\rm{.00 cm}}\).

Substitute \({\rm{9}} \times {\rm{1}}{{\rm{0}}^{\rm{9}}}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}/{{\rm{C}}^{\rm{2}}}\) for \(K\), \[1.00{\rm{ }}\mu {\rm{C}}\] for \(q\), \(3.00{\rm{ cm}}\) for \({r_3}\), \[5.00 cm\] for \({r_5}\), \(8.00{\rm{ cm}}\) for \[{r_8}\], and \(11.00{\rm{ cm}}\) for \({r_{11}}\).

\[\begin{array}{c}\frac{{2 \times \left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left[ {\left( {8.00{\rm{ cm}}} \right) - x} \right]}^2}}} = \frac{{\left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left[ {x - \left( {3.00{\rm{ cm}}} \right)} \right]}^2}}} + \frac{{2 \times \left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left[ {\left( {8.00{\rm{ cm}}} \right) - x} \right]}^2}}} - \frac{{\left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left[ {\left( {11.00{\rm{ cm}}} \right) - x} \right]}^2}}}\\0 = \frac{{\left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left[ {x - \left( {3.00{\rm{ cm}}} \right)} \right]}^2}}} - \frac{{\left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left[ {\left( {11.00{\rm{ cm}}} \right) - x} \right]}^2}}}\\\frac{{\left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left[ {x - \left( {3.00{\rm{ cm}}} \right)} \right]}^2}}} = \frac{{\left( {1.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left[ {\left( {11.00{\rm{ cm}}} \right) - x} \right]}^2}}}\\{\left[ {x - \left( {3.00{\rm{ cm}}} \right)} \right]^2} = {\left[ {\left( {11.00{\rm{ cm}}} \right) - x} \right]^2}\end{array}\]

Taking square root both sides,

\[\begin{array}{c}x - \left( {{\rm{3}}{\rm{.00 cm}}} \right) = \left( {{\rm{11}}{\rm{.00 cm}}} \right) - x\\{\rm{2}}x = \left( {{\rm{11}}{\rm{.00 cm}}} \right) + \left( {{\rm{3}}{\rm{.00 cm}}} \right)\\{\rm{2}}x = {\rm{14}}{\rm{.00 cm}}\\x = {\rm{7}}{\rm{.00 cm}}\end{array}\]\[\]

Hence, the electric field at \(x = {\rm{7}}{\rm{.00 cm}}\) will be same for the \( - 2q\) alone.

(c) Place between 0.00 and 8.00 cm where the electric field is zero

Since, the field is uniform. Hence, there would be not point between \(0.00\) and \(8.00{\rm{ cm}}\) where the electric field is zero.

(d) Electric field rapidly approaches zero

At very large positive or negative values of \(x\), the electric field approaches zero because the distance from the charges approaches infinity.

Since, the net charge in Figure 18.52 (a) is zero. Thus, the electric field approaches zero more rapidly.

(e) Point on right of 11.00 cm where the electric field is zero

Let at \(x\), right of \(11.00{\rm{ cm}}\) electric field be zero.

The electric field at \(x\) is represented as,

Electric field at \(x\)

The electric field at \(x\) is,

\[\begin{array}{c}E = {E_3} - {E_8} + {E_{11}}\\ = \frac{{K\left( q \right)}}{{{{\left( {x - {r_3}} \right)}^2}}} - \frac{{K\left( {2q} \right)}}{{{{\left( {x - {r_8}} \right)}^2}}} + \frac{{K\left( q \right)}}{{{{\left( {x - {r_{11}}} \right)}^2}}}\\ = K\left[ {\frac{{\left( q \right)}}{{{{\left( {x - {r_3}} \right)}^2}}} - \frac{{\left( {2q} \right)}}{{{{\left( {x - {r_8}} \right)}^2}}} + \frac{{\left( q \right)}}{{{{\left( {x - {r_{11}}} \right)}^2}}}} \right]\end{array}\]

Here, \(K\) is the electrostatic force constant, \(q\) is the charge, \(\left( {x - {r_3}} \right)\) is the distance between \(x\) and the charge at \(x = {\rm{3}}{\rm{.00 cm}}\), \(\left( {x - {r_8}} \right)\) is the distance between \(x\) and the charge at \(x = {\rm{8}}{\rm{.00 cm}}\), and \(\left( {x - {r_{11}}} \right)\) is the distance between \(x\) and the charge at \(x = 11.00{\rm{ cm}}\).

Since, the total electric field is zero. Therefore,

\(\begin{array}{c}E = K\left[ {\frac{{\left( q \right)}}{{{{\left( {x - {r_3}} \right)}^2}}} - \frac{{\left( {2q} \right)}}{{{{\left( {x - {r_8}} \right)}^2}}} + \frac{{\left( q \right)}}{{{{\left( {x - {r_{11}}} \right)}^2}}}} \right]\\\frac{E}{{Kq}} = \frac{1}{{{{\left( {x - {r_3}} \right)}^2}}} - \frac{2}{{{{\left( {x - {r_8}} \right)}^2}}} + \frac{1}{{{{\left( {x - {r_{11}}} \right)}^2}}}\end{array}\)

Substitute \[3.00{\rm{ cm}}\] for \({r_3}\), \(5.00{\rm{ cm}}\) for \[{r_5}\], \(8.00{\rm{ cm}}\) for \({r_8}\), and \(11.00{\rm{ cm}}\) for \({r_{11}}\).

\[\frac{E}{{Kq}} = \frac{{\rm{1}}}{{{{\left[ {x - \left( {{\rm{3}}{\rm{.00 cm}}} \right)} \right]}^{\rm{2}}}}} - \frac{{\rm{2}}}{{{{\left[ {x - \left( {{\rm{8}}{\rm{.00 cm}}} \right)} \right]}^{\rm{2}}}}}{\rm{ + }}\frac{{\rm{1}}}{{{{\left[ {x - \left( {{\rm{11}}{\rm{.00 cm}}} \right)} \right]}^{\rm{2}}}}}\]

Plotting this polynomial,

Plot for \(\frac{E}{{Kq}}\) vs \[x\]

From graph, it is clear that the electric field is zero \(x = {\rm{5}}{\rm{.148 cm}}\), \(x = {\rm{9}}{\rm{.745 cm}}\), and \(x = {\rm{30}}{\rm{.607 cm}}\).

Hence, the position right to the point \(x = {\rm{11}}{\rm{.00 cm}}\) where the electric field is zero is \(30.607{\rm{ cm}}\).