# a rod is having a positive charge and a point charge is placed at a distance 'd' from it making...

## Question:

A rod is having a positive charge and a point charge is placed at a distance "D"' from it making angle 45 degrees with one end and 37 degrees with the other. Calculate field.

## Electric Field:

It is the region in which the effect of the charged particle experiences. It varies according to the charge and the distance of the particle. It is also the same as the field lines at a point in the region.

Given Data

• The distance of the charge from the rod is: D.
• The angle of one end of rod from the point charge is:{eq}\alpha = 45^\circ {/eq} .
• The angle of other end of rod from the point charge is: {eq}\beta = 37^\circ {/eq} .

The expression to calculate the angle subtended by the rod on the point charge is,

{eq}\theta = 180^\circ - \alpha - \beta {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} \theta &= 180^\circ - 45^\circ - 37^\circ \\ &= 98^\circ \end{align*} {/eq}

The expression to calculate the distance of one end from the point charge is,

{eq}a = \dfrac{D}{{\sin \alpha }} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} a &= \dfrac{D}{{\sin 45^\circ }}\\ &= \sqrt 2 D \end{align*} {/eq}

The expression to calculate the distance of other end from the point charge is,

{eq}b = \dfrac{D}{{\sin \beta }} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} b &= \dfrac{D}{{\sin 37^\circ }}\\ &= 1.66D \end{align*} {/eq}

The expression to calculate the electric field at the point charge due to one end of the rod is,

{eq}{E_1} = \dfrac{{kQ}}{{{a^2}}} {/eq}

Here, k is the Coulomb's constant and Q is the charge of the rod.

Substitute all the values in the above expression.

{eq}\begin{align*} {E_1} &= \dfrac{{kQ}}{{{{\left( {\sqrt 2 D} \right)}^2}}}\\ &= \dfrac{{kQ}}{{2{D^2}}} \end{align*} {/eq}

The expression to calculate the electric field at the point charge due to other end of the rod is,

{eq}{E_2} = \dfrac{{kQ}}{{{b^2}}} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {E_2} &= \dfrac{{kQ}}{{{{\left( {1.66D} \right)}^2}}}\\ &= \dfrac{{kQ}}{{2.76{D^2}}} \end{align*} {/eq}

The expression to calculate the total electric field at the point charge 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( {\dfrac{{kQ}}{{2{D^2}}}} \right)}^2} + {{\left( {\dfrac{{kQ}}{{2.76{D^2}}}} \right)}^2} + 2\left( {\dfrac{{kQ}}{{2{D^2}}}} \right)\left( {\dfrac{{kQ}}{{2.76{D^2}}}} \right)\cos 98^\circ } \\ &= \sqrt {0.38{{\left( {\dfrac{{kQ}}{{{D^2}}}} \right)}^2} - 0.36{{\left( {\dfrac{{kQ}}{{{D^2}}}} \right)}^2}} \\ &= \sqrt {0.02{{\left( {\dfrac{{kQ}}{{{D^2}}}} \right)}^2}} \\ &= \dfrac{{0.14kQ}}{{{D^2}}} \end{align*} {/eq}

Thus, the total electric field at the point charge is {eq}\dfrac{{0.14kQ}}{{{D^2}}} {/eq}. 