You have a charged gray ring with radius R = 0.25 m. The ring has uniform linear charge density....


You have a charged gray ring with radius {eq}R = 0.25\ m {/eq}. The ring has uniform linear charge density. and the total charge on the ring is {eq}q = 5\ \mu C {/eq}. Hint: The entire length of the ring would be the circle's circumference which is {eq}2 \pi R {/eq}.

a) How many dimensions can the ring be approximated to have?

b) What is the strength of the electric field at a distance {eq}P = 0.5\ m {/eq} along the axis above the middle of the ring as shown in the Diagram?

The following are a few integration identities that may help; within each, {eq}J {/eq} is a constant, and {eq}H {/eq} a variable.

{eq}\displaystyle \int {1 \over 1 + H^2} dH = \tan^{-1} (x)\ \ \ \int {1 \over J^2 + H^2} dH = {1 \over J} \tan^{-1} \left ({H \over J}\right )\ \ \ \int {H \over \sqrt {H + J^3}} dH = \sqrt {H^3 + J^4}\ \ \ \int J dH = JH\\ \displaystyle \int {1 \over \sqrt {H^2 + J^2}} dH = \ln (H) + \sqrt {H^2 + J^2}\ \ \ \int {H \over \sqrt {H^2 + J^2}} dH = \sqrt {H^2 + J^2}\ \ \ \int {1 \over (H^2 + J^2)^{3\over 2} } dH = {H \over J^2 \sqrt {H^2 + J^2}} {/eq}

Electric Field Determination:

The electric field due to a system of charges can be obtained using superposition principle: the total net electric field is the vector sum of all fields due to each individual charge in the system. In the case of continuous charge distribution, the summation shall be replaced by integration.

Answer and Explanation: 1

a) The ring is a one-dimensional structure, since the position of the point on the ring can be uniquely defined by giving only one coordinate, namely an angle with respect to a chosen direction.

b) The electric field is along the x-axis, due to the symmetry of the situation and the fact that the ring is uniformly charged. Therefore, the strength of the field at the point P can be obtained as follows:

{eq}E = \displaystyle \int \limits_0^{2\pi} \dfrac {kq}{2\pi R}\cdot \dfrac {R d\phi}{P^2 + R^2}\cdot \dfrac {P}{\sqrt{P^2 + R^2}} {/eq}


  • {eq}R = 0.25 \ m {/eq} is the radius of the ring;
  • {eq}P = 0.5 \ m {/eq} is the distance to the point of observation.
  • {eq}k = 9\cdot 10^9 \ N\cdot m^2/C^2 {/eq} is the Coulomb's constant;
  • {eq}q = 5\cdot 10^{-6} \ C {/eq} is the total charge on the ring;

Integral is trivial to compute, after which we obtain:

{eq}E = \dfrac {kqP}{(P^2 + R^2)^{\frac {3}{2}}} {/eq}

Computation yields:

{eq}E = \dfrac {9\cdot 10^9 \ N\cdot m^2/C^2 \cdot 5\cdot 10^{-6} \ C \cdot 0.5 \ m}{((0.5 \ m)^2 + (0.25 \ m)^2)^{\frac {3}{2}}} \approx \boxed{1.29\cdot 10^5 \ N/C} {/eq}

Learn more about this topic:

Electric Fields Practice Problems


Chapter 17 / Lesson 8

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