
A ring of radius $R$ has charge $ - Q$ distributed uniformly over it. Calculate the charge that should be placed at the centre of the ring such that the electric field becomes zero at a point on the axis of the ring at distance $R$ from the centre of the ring.
Answer
585.9k+ views
Hint: First let us see what electric field is-
The electric force per unit charge is known as the electric field. The field’s position is taken to be positive of the force on a positive test charge that it will exert. From a positive charge, the electric field is radially outward and radially inward towards a negative point.
The magnitude of the electric field produced by a charging point of magnitude $Q$ , at a distance $r$
from the charging point is given by-
$E = \dfrac{{kQ}}{{{r^2}}}$
where K is a constant
Complete step by step answer:
Given,
A ring of radius $R$ has charge $ - Q$ distributed uniformly over it.
Now we have to calculate the charge that should be placed at the centre of the ring such that the electric field becomes zero at a point on the axis of the ring at distance $R$ from the centre of the ring.
We know that,
$E = \dfrac{{kQ}}{{{r^2}}}$
So, electric field at point $P$
due to charge of ring is-
$E = \dfrac{{kQx}}
{{{{\left( {{R^2} + {x^2}} \right)}^{3/2}}}}$
At $x = R,E = \dfrac{{kQ}}
{{2\sqrt 2 {R^2}}}$ which is towards the centre
So, electric field at $P$ due to charge at centre $\dfrac{{kq}}{{{R^2}}}$
For the net field to be zero at $P$ .
$
\dfrac{{kq}}{{{R^2}}} = \dfrac{{kQ}}{{2\sqrt 2 {R^2}}} \\
q = \dfrac{Q}{{2\sqrt 2 }} \\
$
Additional Information:
The electric field due to a given electrical charge $Q$ is defined as the space around the charge in which another charge $q$ will experience the electrostatic force of attraction or repulsion due to the charge $Q$.
Note:
Here the radius will change with respect to $x$ . So, we have to pay attention to that. Also the resultant field here has to be zero. The electric field can be derived from the potential if the electric potential is known at any point in a region of space.
The electric force per unit charge is known as the electric field. The field’s position is taken to be positive of the force on a positive test charge that it will exert. From a positive charge, the electric field is radially outward and radially inward towards a negative point.
The magnitude of the electric field produced by a charging point of magnitude $Q$ , at a distance $r$
from the charging point is given by-
$E = \dfrac{{kQ}}{{{r^2}}}$
where K is a constant
Complete step by step answer:
Given,
A ring of radius $R$ has charge $ - Q$ distributed uniformly over it.
Now we have to calculate the charge that should be placed at the centre of the ring such that the electric field becomes zero at a point on the axis of the ring at distance $R$ from the centre of the ring.
We know that,
$E = \dfrac{{kQ}}{{{r^2}}}$
So, electric field at point $P$
due to charge of ring is-
$E = \dfrac{{kQx}}
{{{{\left( {{R^2} + {x^2}} \right)}^{3/2}}}}$
At $x = R,E = \dfrac{{kQ}}
{{2\sqrt 2 {R^2}}}$ which is towards the centre
So, electric field at $P$ due to charge at centre $\dfrac{{kq}}{{{R^2}}}$
For the net field to be zero at $P$ .
$
\dfrac{{kq}}{{{R^2}}} = \dfrac{{kQ}}{{2\sqrt 2 {R^2}}} \\
q = \dfrac{Q}{{2\sqrt 2 }} \\
$
Additional Information:
The electric field due to a given electrical charge $Q$ is defined as the space around the charge in which another charge $q$ will experience the electrostatic force of attraction or repulsion due to the charge $Q$.
Note:
Here the radius will change with respect to $x$ . So, we have to pay attention to that. Also the resultant field here has to be zero. The electric field can be derived from the potential if the electric potential is known at any point in a region of space.
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