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A straight, nonconducting plastic wire, 8.50-cm-long, carries a charge density of 125nC/m,...

Question:

A straight, nonconducting plastic wire, 8.50-cm-long, carries a charge density of 125nC/m, distributed uniformly along its length. It is lying on a horizontal tabletop.

A) Find the magnitude and direction of the electric field this wire produces at a point, 5.50cm directly above its midpoint.

B) If the wire is now bent into a circle, lying flat on the table, find the magnitude and direction of the electric field it produces at a point, 5.50 cm directly above its center.

Current:

Current is the rate of time at which charge flows through some space, the direction of the free positive charges provides the direction of the current. The standard unit of current is ampere.

Answer and Explanation:


Given data


  • Length of the non-conducting wire, {eq}L = 8.50\;{\rm{cm}} = 0.0850\;{\rm{m}} {/eq}
  • Charge density, {eq}\lambda = 125\;{\rm{nC/m}} = 125 \times {10^{ - 9}}\;{\rm{nC/m}} {/eq}


(a) Magnitude and direction of the electric field at a point {eq}x = 5.50\;{\rm{cm}} {/eq} above from the mid-point


Expression of the electric field at mid point in the case of long wire can be given as


{eq}E = \dfrac{{k\lambda L}}{{x\sqrt {{x^2} + {{\left( {\dfrac{L}{2}} \right)}^2}} }} {/eq}


Here {eq}k {/eq} is the coulombs constant, {eq}\lambda {/eq} is the charge density, {eq}x {/eq} is the distance from mid point of the wire and {eq}L {/eq} is the length of the wire


Take {eq}k = 9 \times {10^9}\;{\rm{N}}{{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2} {/eq}


Now substitute the values in the expression of the electric field, we get


{eq}\begin{align*} E &= \dfrac{{9 \times {{10}^9} \times 125 \times {{10}^{ - 9}} \times 0.0850}}{{0.0550\sqrt {{{0.0550}^2} + {{\left( {\dfrac{{0.0850}}{2}} \right)}^2}} }}\\ E &= \dfrac{{95.625}}{{3.8228 \times {{10}^{ - 3}}}}\\ E &= 2.5014 \times {10^4}\;{\rm{N/C}} \end{align*} {/eq}


Thus, magnitude of the electric field is {eq}2.5014 \times {10^4}\;{\rm{N/C}} {/eq} and its direction is upward because it has only X component.


(b) Expression of the electric field in the case circle can be given as


{eq}E = \dfrac{{k\lambda Lx}}{{{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}} {/eq}


Here a is the radius of the circle and can be given as


{eq}\begin{align*} L &= 2\pi a\\ a &= \left( {\dfrac{L}{{2\pi }}} \right) \end{align*} {/eq}


By substituting the values in electric field expression, we get


{eq}\begin{align*} E &= \dfrac{{9 \times {{10}^9} \times 125 \times {{10}^{ - 9}} \times 0.0850 \times 0.0550}}{{{{\left( {{{0.0550}^2} + {{\left( {\dfrac{{0.0850}}{{2\pi }}} \right)}^2}} \right)}^{3/2}}}}\\ &= \dfrac{{5.2593}}{{1.8169 \times {{10}^{ - 4}}}}\\ &= 2.8946 \times {10^4}\;{\rm{N/C}} \end{align*} {/eq}


Thus, magnitude of the electric field above the center is {eq}2.8946 \times {10^4}\;{\rm{N/C}} {/eq} and its direction is upward because it has only X component.



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