Two charged particles are located on x axis: q_1 = -3.20 \times 10^{-19}\ C at x = -3\ m and q_2...

Question:

Two charged particles are located on x axis: {eq}q_1 = -3.20 \times 10^{-19}\ C {/eq} at {eq}x = -3\ m {/eq} and {eq}q_2 = 6 \times 10^{-19}\ C {/eq} at {eq}x = +3\ m {/eq}. What are a) the magnitude and b) direction (relative to the positive direction of x axis) of the net electric field at point P at {eq}y = 4.00\ m {/eq}?

Electric Field:

The vicinity of the charge in which the effect of the charge experiences is called the electric field. It varies according to the charge and distance of the particle. It is the same as the field lines at the point.

Answer and Explanation:


Given Data

  • The charge of the first particle is: {eq}{q_1} = - 3.20 \times {10^{ - 19}}\;{\rm{C}} {/eq}.
  • The charge of the second particle is: {eq}{q_2} = 6 \times {10^{ - 19}}\;{\rm{C}} {/eq}.
  • The position of the first point is: {eq}{x_1} = - 3\;{\rm{m}} {/eq}.
  • The position of the second point is: {eq}{x_2} = 3\;{\rm{m}} {/eq}.
  • The position of the point P is: {eq}y = 4\;{\rm{m}} {/eq}.


The expression to calculate the distance of the point P from the first particle is,

{eq}{r_1} = \sqrt {x_1^2 + y} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {r_1} &= \sqrt {{{\left( { - 3\;{\rm{m}}} \right)}^2} + {{\left( {4\;{\rm{m}}} \right)}^2}} \\ &= 5\;{\rm{m}} \end{align*} {/eq}


The expression to calculate the distance of the point P from the second particle is,

{eq}{r_2} = \sqrt {x_2^2 + y} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {r_2} &= \sqrt {{{\left( {3\;{\rm{m}}} \right)}^2} + {{\left( {4\;{\rm{m}}} \right)}^2}} \\ &= 5\;{\rm{m}} \end{align*} {/eq}


The expression to calculate the electric field at the point P due to first particle is,

{eq}{E_1} = \dfrac{{k{q_1}}}{{r_1^2}} {/eq}

Here, k is the Coulomb's constant and its value is 9 \times {10^9}\;{\rm{N} \cdot {eq}{{\rm{m}}^2}/{{\rm{C}}^2} {/eq}.

Substitute all the values in the above expression.

{eq}\begin{align*} {E_1} &= \dfrac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( { - 3.20 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}{{{{\left( {5\;{\rm{m}}} \right)}^2}}}\\ &= - 1.152 \times {10^{ - 10}}\;{\rm{N}}/{\rm{C}}\; \end{align*} {/eq}


The expression to calculate the electric field at the point P due to second particle is,

{eq}{E_2} = \dfrac{{k{q_2}}}{{r_2^2}} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {E_2} &= \dfrac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}/{{\rm{C}}^2}} \right)\left( {6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}{{{{\left( {5\;{\rm{m}}} \right)}^2}}}\\ &= 2.16 \times {10^{ - 10}}\;{\rm{N}}/{\rm{C}}\; \end{align*} {/eq}


The expression to calculate the angle between the electric fields at the point P is,

{eq}\tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{{{x_2}}}{y} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} \tan \left( {\dfrac{\theta }{2}} \right) &= \dfrac{{3\;{\rm{m}}}}{{4\;{\rm{m}}}}\\ \dfrac{\theta }{2}& = 36.87^\circ \\ \theta &= 73.74^\circ \end{align*} {/eq}


(a)

The expression to calculate the magnitude of the electric field at the point P is,

{eq}E = \sqrt {E_1^2 + E_2^2 + 2{E_1}{E_2}\cos \theta } {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} E &= \sqrt {{{\left( { - 1.152 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}\;} \right)}^2} + {{\left( {2.16 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}\;} \right)}^2} + 2\left( { - 1.152 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}\;} \right)\left( {2.16 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}\;} \right)\cos 73.74^\circ } \\ &= 2.145 \times {10^{ - 10}}\;{\rm{N}}/{\rm{C}} \end{align*} {/eq}

Thus, the magnitude of the electric field at the point P is {eq}2.145 \times {10^{ - 10}}\;{\rm{N}}/{\rm{C}} {/eq}.



(b)

The expression to calculate the direction of electric field at the point P is,

{eq}\tan \alpha = \dfrac{{{E_2}\sin \theta }}{{{E_1} + {E_2}\cos \theta }} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} \tan \alpha &= \dfrac{{\left( {2.16 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}\;} \right)\left( {\sin 73.74^\circ } \right)}}{{ - 1.152 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}\; + \left( {2.16 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}\;} \right)\left( {\cos 73.74^\circ } \right)}}\\ \tan \alpha & = \dfrac{{2.074 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}}}{{ - 0.547 \times {{10}^{ - 10}}\;{\rm{N}}/{\rm{C}}}}\\ \alpha &= - 75.23^\circ \end{align*} {/eq}


Thus, the direction of electric field at the point P is {eq}- 75.23^\circ {/eq} .


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