Question

A uniformly charged rod of length 4cm and linear charge density $\lambda = 30{\rm{ }}\mu {{\rm{C}} {\left/{\vphantom {{\rm{C}} {\rm{m}}}} \right.} {\rm{m}}}$ is placed as shown in figure. Calculate the x-component of electric field at point P

Answer 1

We will use the expression for electric field in x-direction of the rod which gives us the relation between linear charge density, vertical distance between point P and rod, Coulomb’s constant and angles made by electric field with the vertical.

Complete step by step answer:
Given:
The length of rod is $l = 4{\rm{ cm}} \times \left( {\dfrac{{\rm{m}}}{{100{\rm{ cm}}}}} \right) = 0.04{\rm{ m}}$.
The distance between rod and point P is $r = 3{\rm{ cm}} = 3{\rm{ cm}} \times \left( {\dfrac{{\rm{m}}}{{100{\rm{ cm}}}}} \right) = 0.03{\rm{ m}}$.
The linear charge density is $\lambda = 30{\rm{ }}\mu {{\rm{C}} {\left/ {\vphantom {{\rm{C}} {\rm{m}}}} \right. } {\rm{m}}}$.
We have to calculate the x-component of electric point P.
Let us write the general expression for the electric field in the x-direction of rod.
${E_x} = \dfrac{{k\lambda }}{r}\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$……(1)
Here k is the Coulomb’s constant for air, ${\theta _1}$ is the angle at point P and
${\theta _2}$ is the angle at another point Q if it exists in negative y-direction in a similar fashion as that of P.
We know that the value of Coulomb’s constant is given as:

{\vphantom {{{{\rm{m}}^2}} {{{\rm{C}}^2}}}} \right. } {{{\rm{C}}^2}}}$$ Let us redraw the above figure as below:  From the given figure we can conclude that point Q is not present in our problem so we can substitute zero for $${\theta _2}$$. We can use the tangent of angle $$\theta $$ to find its value from the above diagram.

\tan \theta = \left( {\dfrac{{{\rm{4 cm}}}}{{3{\rm{ cm}}}}} \right)\\
= \dfrac{4}{3}

$$Taking the inverse of tangent on both sides of the above expression, we get:$$

\theta = {\tan ^{ - 1}}\left( {\dfrac{4}{3}} \right)\\
= 53.13^\circ

On substituting $$9 \times {10^9}{\rm{ N}} \cdot {{{{\rm{m}}^2}} {\left/ {\vphantom {{{{\rm{m}}^2}} {{{\rm{C}}^2}}}} \right. } {{{\rm{C}}^2}}}$$ for k, $$30{\rm{ }}\mu {{\rm{C}} {\left/ {\vphantom {{\rm{C}} {\rm{m}}}} \right. } {\rm{m}}}$$ for $$\lambda $$, $$0.03{\rm{ m}}$$ for r $$53.13^\circ $$ for $${\theta _1}$$ and $$0^\circ $$ for $${\theta _2}$$ in equation (1), we get:

{E_x} = \dfrac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{{{\rm{m}}^2}} {\left/
{\vphantom {{{{\rm{m}}^2}} {{{\rm{C}}^2}}}} \right.
} {{{\rm{C}}^2}}}} \right)\left( {30{\rm{ }}\mu {{\rm{C}} {\left/
{\vphantom {{\rm{C}} {\rm{m}}}} \right.
} {\rm{m}}}} \right)}}{{0.03{\rm{ m}}}}\left( {\cos 53.13^\circ - \cos 0^\circ } \right)\\
= \dfrac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{{{\rm{m}}^2}} {\left/
{\vphantom {{{{\rm{m}}^2}} {{{\rm{C}}^2}}}} \right.
} {{{\rm{C}}^2}}}} \right)\left( {30 \times {{10}^{ - 6}}{{\rm{C}} {\left/
{\vphantom {{\rm{C}} {\rm{m}}}} \right.
} {\rm{m}}}} \right)}}{{0.03{\rm{ m}}}}\left( {\cos 53.13^\circ - \cos 0^\circ } \right)\\
= - 36 \times {10^5}{\rm{ }}{{\rm{N}} {\left/
{\vphantom {{\rm{N}} {\rm{C}}}} \right.
} {\rm{C}}}

Therefore, we can write that the value of electric field in x-direction is $$ - 36 \times {10^5}{\rm{ }}{{\rm{N}} {\left/{\vphantom {{\rm{N}} {\rm{C}}}} \right.} {\rm{C}}}$$. **Note:** The negative value of electric field signifies that the direction of electric field is opposite to our consideration. Also, we should not be confused with the value of $${\theta _2}$$ as it is not present in our problem so we can substitute it equal to zero.