Two 10-cm-diameter charged rings face each other, 20.0 cm apart. Both rings are charged to +...

Question:

Two 10-cm-diameter charged rings face each other, 20.0 cm apart. Both rings are charged to + 30.0nC. What is the electric field strength at the center of the left ring?

Electric field:

A field originated by the charge of electrical where electric charge exerted force on each other is known as an electric field. It is a vector quantity that shows the direction of the field of electric.

Answer and Explanation:

  • The diameter of the ring is {eq}d = 10\,{\rm{cm}} = 0.1\,{\rm{m}} {/eq}
  • The distance between two charged rings is {eq}x = 20\,{\rm{cm}} = 0.2\,{\rm{m}} {/eq}
  • Both rings are charged is {eq}Q = 30\,{\rm{nC}} = 30 \times {10^{ - 9}}{\rm{C}} {/eq}


The expression for the radius of the ring is

{eq}a = \dfrac{d}{2} {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} a &= \dfrac{{0.1}}{2}\\ &= 0.05\,{\rm{m}} \end{align*} {/eq}

The expression for the electric field at the centre of the left ring dur to the right ring is

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

  • Here {eq}k = 9 \times {10^9} {/eq} is the coloumb constant.

Substituting the values in the above expression as,

{eq}\begin{align*} E &= \dfrac{{\left( {9 \times {{10}^9}} \right)\left( {30 \times {{10}^{ - 9}}} \right)\left( {0.2} \right)}}{{{{\left( {{{\left( {0.2} \right)}^2} + {{\left( {0.5} \right)}^2}} \right)}^{\dfrac{3}{2}}}}}\\ & = \frac{{270 \times 0.2}}{{{{\left( {\left( {0.04} \right) + \left( {0.025} \right)} \right)}^{\frac{3}{2}}}}}\\ &= 6164.38\,{\rm{N/C}} \end{align*} {/eq}

Thus the electric field strength at the centre of the left ring is {eq}E = 6164.38\,{\rm{N/C}} {/eq}


Learn more about this topic:

Loading...
Electric Fields Practice Problems

from Physics 101: Intro to Physics

Chapter 17 / Lesson 8
3.3K

Related to this Question

Explore our homework questions and answers library