Courses
Courses for Kids
Free study material
Offline Centres
More
Last updated date: 25th Nov 2023
Total views: 343.2k
Views today: 3.43k

# A charge $q = 10\,{\text{mC}}$ is distributed uniformly over the circumference of a ring of radius $3\,{\text{m}}$ placed on x-y plane with its centre at origin. Find the electric potential at a point $P\left( {0,0,4\,{\text{m}}} \right)$A. $18\,{\text{V}}$B. $1.8 \times {10^2}\,{\text{V}}$C. $1.8 \times {10^3}\,{\text{V}}$D. $1.8 \times {10^7}\,{\text{V}}$

Verified
343.2k+ views
Hint: First of all, we will find the shortest distance between the point and the charge. Then we will apply the formula and substitute the required values and manipulate accordingly to find the electric potential.

Complete step by step answer:In the given question, we are supplied with the following data:
The charge present which is uniformly distributed over the circumference of a ring is $10\,{\text{mC}}$ .
The radius of the ring is $3\,{\text{m}}$ .
We are asked to find the electric potential at a point $P\left( {0,0,4\,{\text{m}}} \right)$ .

To proceed the numerical, we will convert the unit of charge to S.I units:
We know,
$1\,{\text{mC}} = 1 \times {10^{ - 3}}\,{\text{C}}$
So, we have:
$10\,{\text{mC}} = 10 \times {10^{ - 3}}\,{\text{C}}$

Now, we need to find the electric potential due to the given charge, at a point which is at a distance of $3\,{\text{m}}$ from the charge.
So, we calculate the shortest distance between them by using the Pythagoras theorem, as given below:
$r = \sqrt {{3^2} + {4^2}} \\ r = \sqrt {9 + 16} \\ r = \sqrt {25} \\ r = 5\,{\text{m}} \\$
Therefore, the shortest distance has come out to be $5\,{\text{m}}$ .
Now, to find the electric potential we apply the formula, as given below:
$V = \dfrac{{Kq}}{r}$ …… (1)
Where,
$V$ indicates electric potential at a point.
$K$ indicates Coulomb’s constant.
$q$ indicates point charge.
$r$ indicates the shortest distance between the circumference and the given point.
Substituting the required values in the equation (1), we get:

$V = \dfrac{{Kq}}{r} \\ V = \dfrac{{9 \times {{10}^9} \times 10 \times {{10}^{ - 3}}}}{5} \\ V = \dfrac{{90 \times {{10}^6}}}{5} \\ V = 1.8 \times {10^7}\,{\text{V}} \\$
Hence, electric potential at a point $P\left( {0,0,4\,{\text{m}}} \right)$ is $1.8 \times {10^7}\,{\text{V}}$ .

The correct option is D.

Note:While solving the numerical, many students tend to make mistake by taking the distance as $3\,{\text{m}}$ into account, however it is $5\,{\text{m}}$ , as it is located in space. So, we need to find the shortest distance between the point charge and the point mentioned in the question.