A positively charged ring is in the y-z plane | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The electric potential $V$ at a distance $x$ from the center of a charged ring of radius $a$ and charge $q$ is given by $V = \frac{kq}{\sqrt{x^2 +

Vedclass · Accepted Answer

Increases continuously

Vedclass · Answer

(A) The electric potential $V$ at a distance $x$ from the center of a charged ring of radius $a$ and charge $q$ is given by $V = \frac{kq}{\sqrt{x^2 + a^2}}$.At the origin $(x=0)$,the potential is $V_{center} = \frac{kq}{a}$.As the test charge $q_0$ moves along the $x$-axis away from the origin,the distance $\sqrt{x^2 + a^2}$ increases,which means the potential $V$ decreases.The potential energy $U$ of the test charge is $U = q_0 V$. Since $V$ decreases as $x$ increases,the potential energy $U$ also decreases.According to the law of conservation of energy,the total mechanical energy remains constant. As the potential energy $U$ decreases,the kinetic energy $K$ must increase to compensate.Therefore,the speed of the test charge increases continuously as it moves away from the origin.

$A$ positively charged ring is in the $y-z$ plane with its centre at the origin. $A$ positive test charge $q_0$,held at the origin,is released along the $x$-axis. Then its speed

Similar Questions

$A$ thin conducting spherical shell of radius $R$ has a charge $q$. Another charge $Q$ is placed at the center of the shell. The electrostatic potential at a point $P$ at a distance $R/2$ from the center of the shell is:

Four point charges are placed at the vertices of a square of side $2.8 \,m$ as shown in the figure. Find the electric potential at the centre of the square. (in $\,V$)

Two equal point charges are fixed at $x = -a$ and $x = +a$ on the $x$-axis. Another point charge $Q$ is placed at the origin. The change in the electrical potential energy of $Q$,when it is displaced by a small distance $x$ along the $x$-axis,is approximately proportional to

If a unit positive charge is shifted from a region of low potential to a region of high potential,then the electric potential energy of the system

Three particles,each having a charge of $10\,\mu C$,are placed at the corners of an equilateral triangle of side $10\,cm$. The electrostatic potential energy of the system is.....$J$ (Given $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\,N\cdot m^2/C^2$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module