91Ó°ÊÓ

The space around the charge in which another charge experiences some electrostatic force of attraction or repulsion 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 magnitude of the charge and\(r\)is the distance of the point of consideration from the charge.

The electric field is a vector quantity. For a system of charges, the electric field at a point equals the vector sum of all the electric field acting at that point.

Let the electric field be zero at point x from the \(25.0{\rm{ }}\mu C\). The electric field at x is represented as,

Here,\({E_1}\)is the electric field due to the charge\(25.0{\rm{ \mu C}}\),\({E_2}\)is the electric field due to the charge\(45.0{\rm{ \mu C}}\), and\(r\)is the distance between\(25.0{\rm{ \mu C}}\)and\(45.0{\rm{ \mu C}}\).

The electric field due to the charge\(25.0{\rm{ \mu C}}\)is,

\({E_1} = \frac{{K\left( {25.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{x^2}}}\)

The electric field due to the charge\(45.0{\rm{ \mu C}}\)is,

\({E_2} = \frac{{K\left( {45.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {r - x} \right)}^2}}}\)

The net electric field at point P is,

\(\begin{array}{c}E = {E_1} - {E_2}\\ = \frac{{K\left( {25.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{x^2}}} - \frac{{K\left( {45.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {r - x} \right)}^2}}}\end{array}\)

Since, the net electric field at point P is zero i.e.,\(E = 0\). Therefore,

\(\begin{array}{c}0 = \frac{{K\left( {25.0{\rm{ \mu C}}} \right)}}{{{x^2}}} - \frac{{K\left( {45.0{\rm{ \mu C}}} \right)}}{{{{\left( {r - x} \right)}^2}}}\\\frac{{K\left( {25.0{\rm{ \mu C}}} \right)}}{{{x^2}}} = \frac{{K\left( {45.0{\rm{ \mu C}}} \right)}}{{{{\left( {r - x} \right)}^2}}}\\{\left( {\frac{{r - x}}{x}} \right)^2} = \frac{{45}}{{25}}\end{array}\)

Therefore, the expression for\(x\)(distance where electric field is zero from\(25.0{\rm{ \mu C}}\)) is,

\(\begin{array}{c}\frac{r}{x} - 1 = \sqrt {\frac{{45}}{{25}}} \\\frac{r}{x} = \sqrt {\frac{{45}}{{25}}} + 1\\x = \frac{r}{{\sqrt {\frac{{45}}{{25}}} + 1}}\end{array}\)

Substitute\(0.500{\rm{ m}}\)for\(r\),

\(\begin{array}{c}x = \frac{{0.500{\rm{ m}}}}{{\sqrt {\frac{{45}}{{25}}} + 1}}\\ = 0.214{\rm{ m}}\end{array}\)

Hence, the electric field will be zero at \(0.214{\rm{ m}}\) from the \(25.0{\rm{ \mu C}}\).

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

Here,\({E_1}\)is the electric field due to the charge\(25.0{\rm{ \mu C}}\),\({E_2}\)is the electric field due to the charge\(45.0{\rm{ \mu C}}\), and\(r\)is the distance between\(25.0{\rm{ \mu C}}\)and\(45.0{\rm{ \mu C}}\).

The electric field due to the charge\(25.0{\rm{ \mu C}}\)is,

\({E_1} = \frac{{K\left( {25.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {\frac{r}{2}} \right)}^2}}}\)

The electric field due to the charge\(45.0{\rm{ \mu C}}\)is,

\({E_2} = \frac{{K\left( {45.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {\frac{r}{2}} \right)}^2}}}\)

The net electric field at point P is,

Substitute\(9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\)for\(K\), and\(0.500\,{\rm{m}}\)for r,

\(\begin{array}{c}E = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( { - 20.0{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {\frac{{0.500}}{2}} \right)}^2}}}\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \left( { - 20.0{\rm{ }}\mu {\rm{C}}} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)}}{{{{\left( {\frac{{0.500}}{2}} \right)}^2}}}\\ = - 2.88 \times {10^6}{\rm{ N}}/{\rm{C}}\end{array}\)

Here, the negative sign indicates that the electric field is directed away from the\(45.0{\rm{ \mu C}}\)charge.

Hence, the electric field in the half way between the charges is\(2.88 \times {10^6}{\rm{ N}}/{\rm{C}}\)directed away from the\(45.0{\rm{ \mu C}}\).