91Ó°ÊÓ

A system consists of two equal but opposite nature of charge separated by some distance is called electric dipole.

Consider two metal balls having charges \( + 5{\rm{ C}}\) and \( - 5{\rm{ C}}\) are separated by \(6{\rm{ m}}\). (a) Find the magnitude of electric field between half way of the charges. (b) Find the magnitude of electric field at any point \(4{\rm{ m}}\) on the equatorial line. (c) Find the magnitude of electric field either on both side at distance of \(15{\rm{ m}}\) on the line joining both charges.

The electric field in the half way of the charges is represented as,

Here, \({E_1}\) is the electric field due to \( + 5{\rm{ C}}\), \({E_2}\) is the electric field due to \( - 5{\rm{ C}}\), and \(r\) is the separation between \( + 5{\rm{ C}}\) and \( - 5{\rm{ C}}\).

The electric field due to \( + 5{\rm{ C}}\) is,

\({E_1} = \frac{{K\left| {5{\rm{ C}}} \right|}}{{{{\left( {\frac{r}{2}} \right)}^2}}}\)

Here, \(K\) is the electrostatic force constant.

Substitute \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\) for \(K\), and \(6{\rm{ m}}\) for \(r\),

\(\begin{array}{c}{E_1} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left| {5{\rm{ C}}} \right|}}{{{{\left( {\frac{{6{\rm{ m}}}}{2}} \right)}^2}}}\\ = 5 \times {10^9}{\rm{ N}}/{\rm{C}}\end{array}\)

The electric field due to \( - 5{\rm{ C}}\) is,

\({E_2} = \frac{{K\left| { - 5{\rm{ C}}} \right|}}{{{{\left( {\frac{r}{2}} \right)}^2}}}\)

Here, \(K\) is the electrostatic force constant.

Substitute \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\) for \(K\), and \(6{\rm{ m}}\) for \(r\),

\(\begin{array}{c}{E_2} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left| { - 5{\rm{ C}}} \right|}}{{{{\left( {\frac{{6{\rm{ m}}}}{2}} \right)}^2}}}\\ = 5 \times {10^9}{\rm{ N}}/{\rm{C}}\end{array}\)

At the center, the net electric field is,

\(E = {E_1} + {E_2}\)

Substitute \(5 \times {10^9}{\rm{ N}}/{\rm{C}}\)for\({E_1}\)and\(5 \times {10^9}{\rm{ N}}/{\rm{C}}\)for\({E_2}\),

\(\begin{array}{c}E = \left( {5 \times {{10}^9}{\rm{ N}}/{\rm{C}}} \right) + \left( {5 \times {{10}^9}{\rm{ N}}/{\rm{C}}} \right)\\ = {10^{10}}{\rm{ N}}/{\rm{C}}\end{array}\)

Hence, the electric field in the halfway of the charges is \({10^{10}}{\rm{ N}}/{\rm{C}}\).

The electric field on the equatorial line is represented as,

Here, \({E_1}\) is the electric field due to \( + 5{\rm{ C}}\), \({E_2}\) is the electric field due to \( - 5{\rm{ C}}\), and \(r\) is the separation between \( + 5{\rm{ C}}\) and \( - 5{\rm{ C}}\), \(x\) is the distance of the point from the equatorial line.

The distance of the point from the charge is,

\(d = \sqrt {{{\left( {\frac{r}{2}} \right)}^2} + {x^2}} \)

Substitute \(6{\rm{ m}}\) for \(r\), and \(4{\rm{ m}}\) for \(x\),

\(\begin{array}{c}d = \sqrt {{{\left( {\frac{{6{\rm{ m}}}}{2}} \right)}^2} + {{\left( {4{\rm{ m}}} \right)}^2}} \\ = 5{\rm{ m}}\end{array}\)

The electric field due to \( + 5{\rm{ C}}\) is,

\({E_1} = \frac{{K\left| {5{\rm{ C}}} \right|}}{{{d^2}}}\)

Here, \(K\) is the electrostatic force constant.

Substitute \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\) for \(K\), and \({\rm{5 m}}\) for \(d\),

\(\begin{array}{c}{E_1} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left| {5{\rm{ C}}} \right|}}{{{{\left( {5{\rm{ m}}} \right)}^2}}}\\ = 1.8 \times {10^9}{\rm{ N}}/{\rm{C}}\end{array}\)

The electric field due to \( - 5{\rm{ C}}\) is,

\({E_2} = \frac{{K\left| { - 5{\rm{ C}}} \right|}}{{{d^2}}}\)

Here, \(K\) is the electrostatic force constant.

Substitute \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\) for \(K\), and \({\rm{5 m}}\) for \(d\),

\(\begin{array}{c}{E_2} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left| { - 5{\rm{ C}}} \right|}}{{{{\left( {5{\rm{ m}}} \right)}^2}}}\\ = 1.8 \times {10^9}{\rm{ N}}/{\rm{C}}\end{array}\)

From geometry,

\(\cos \theta = \frac{{r/2}}{d}\)

Substitute \(6{\rm{ m}}\) for \(r\), and \(5{\rm{ m}}\) for \(d\),

\(\begin{array}{c}\cos \theta = \frac{{\left( {6{\rm{ m}}/2} \right)}}{{5{\rm{ m}}}}\\ = 0.6\end{array}\)

Due to symmetry of the system, the horizontal component of the field cancels each other. Therefore, the net electric field at the point on equatorial line is,

\(\begin{array}{c}E = {E_1}\cos \theta + {E_2}\cos \theta \\ = \left( {{E_1} + {E_2}} \right)\cos \theta \end{array}\)

Substitute \(1.8 \times {10^9}{\rm{ N}}/{\rm{C}}\) for \({E_1}\), \(1.8 \times {10^9}{\rm{ N}}/{\rm{C}}\) for \({E_2}\), and \(0.6\) for \(\cos \theta \),

\(\begin{array}{c}E = \left[ {\left( {1.8 \times {{10}^9}{\rm{ N}}/{\rm{C}}} \right) + \left( {1.8 \times {{10}^9}{\rm{ N}}/{\rm{C}}} \right)} \right] \times 0.6\\ = 2.16 \times {10^9}{\rm{ N}}/{\rm{C}}\end{array}\)

Hence, the electric field strength at a distance the \(4{\rm{ m}}\) on the equatorial line is \(2.16 \times {10^9}{\rm{ N}}/{\rm{C}}\).

The electric field is represented as,

Here, \({E_1}\) is the electric field due to \( + 5{\rm{ C}}\), \({E_2}\) is the electric field due to \( - 5{\rm{ C}}\), and \(r\) is the separation between \( + 5{\rm{ C}}\) and \( - 5{\rm{ C}}\), \(d\) is the charge.

The electric field due to \( + 5{\rm{ C}}\) is,

\({E_1} = \frac{{K\left| {5{\rm{ C}}} \right|}}{{{{\left( {r + d} \right)}^2}}}\)

Here, \(K\) is the electrostatic force constant.

Substitute \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\) for \(K\), \(6{\rm{ m}}\) for \(r\), and \({\rm{15 m}}\) for \(d\),

\(\begin{array}{c}{E_1} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left| {5{\rm{ C}}} \right|}}{{{{\left[ {\left( {{\rm{6 m}}} \right) + \left( {15{\rm{ m}}} \right)} \right]}^2}}}\\ = 1.02 \times {10^8}{\rm{ N}}/{\rm{C}}\end{array}\)

The electric field due to \( - 5{\rm{ C}}\) is,

\({E_2} = \frac{{K\left| { - 5{\rm{ C}}} \right|}}{{{d^2}}}\)

Substitute \(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}\) for \(K\), and \({\rm{15 m}}\) for \(d\),

\(\begin{array}{c}{E_2} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right) \times \left| { - 5{\rm{ C}}} \right|}}{{{{\left( {15{\rm{ m}}} \right)}^2}}}\\ = 2 \times {10^8}{\rm{ N}}/{\rm{C}}\end{array}\)

The magnitude of electric field at distance of \(15{\rm{ m}}\) is,

\(E = \left| {{E_1} - {E_2}} \right|\)

Substitute \(1.02 \times {10^8}{\rm{ N}}/{\rm{C}}\) for \({E_1}\) and \(2 \times {10^8}{\rm{ N}}/{\rm{C}}\) for \({E_2}\),

\(\begin{array}{c}E = \left| {\left( {1.02 \times {{10}^8}{\rm{ N}}/{\rm{C}}} \right) - \left( {2 \times {{10}^8}{\rm{ N}}/{\rm{C}}} \right)} \right|\\ = 9.8 \times {10^7}{\rm{ N}}/{\rm{C}}\end{array}\)

When the test charge is on the right side of the \( + 5{\rm{ C}}\), due to the symmetry of the system the electric field is \(9.8 \times {10^7}{\rm{ N}}/{\rm{C}}\).

Hence, the magnitude of the electric field \(9.8 \times {10^7}{\rm{ N}}/{\rm{C}}\).