Question

A charged ring of radius 0.5 m has a 0.002 π m gap. If the ring carries a charge of +1 C, the electric field at the centre is:

Answer 1

Hint: We all know that for the body to be electrically neutral, there must be an excess of positive charges or negative charges. If there would be no hole, then the positive charge would be in quantity, and then there is a generation of an electric field.

Complete step by step solution:
Given:
The length of the gap is $l = 0.002\pi \;{\rm{m}}$.
The charge carried by the ring is $Q = + 1\;{\rm{C}}$.
The radius of the ring is $R = 0.5\;{\rm{m}}$.
We know that to find the charge in the hole, we can use the formula,
\[q = \dfrac{{Ql}}{{2\pi R}}\]
Here, q is the charge on the gap.
We will now substitute $l = 0.002\pi \;{\rm{m}}$ , $Q = + 1\;{\rm{C}}$ $R = 0.5\;{\rm{m}}$ in the above equation, to find the value of q.
\[\begin{array}{l}
q = \dfrac{{\left( { + 1\;{\rm{C}}} \right)\left( {0.002\pi \;{\rm{m}}} \right)}}{{2\pi \left( {0.5\;{\rm{m}}} \right)}}\\
 = 2 \times {10^{ - 3}}\;{\rm{C}}
\end{array}\]
We will now write the formula to find the electric field at the center due to the given ring.
$E = \dfrac{{kq}}{{{R^2}}}$
Here E is the electric field, k is the coulomb’s constant, and its value is $9 \times {10^9}\dfrac{{{\rm{N}}{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , r is the distance between ring and center.
We will now substitute $q = 2 \times {10^{ - 3}}\;{\rm{C}}$ $R = 0.5\;{\rm{m}}$ and $k = 9 \times {10^9}\dfrac{{{\rm{N}}{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ to find the value of E.
$\begin{array}{l}
E = \dfrac{{\left( {9 \times {{10}^9}\dfrac{{{\rm{N}}{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}} \right)\left( {2 \times {{10}^{ - 3}}\;{\rm{C}}} \right)}}{{{{\left( {0.5\;{\rm{m}}} \right)}^2}}}\\
 = 7.2 \times {10^7}\;\dfrac{{\rm{N}}}{{\rm{C}}}
\end{array}$

Therefore, the electric field at the center is $E = 7.2 \times {10^7}\;\dfrac{{\rm{N}}}{{\rm{C}}}$.

Note: Here in this problem, we are finding the electric field, and the electric field is directed at a point where there is a positive test charge and not at the end of a negative charge. Here in this problem, the electric field is from the center to the ring.