A particle A has charge + q and a particle B has charge + ,4q with each of them having same mass m .

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2. Electric Potential and Capacitance

easy

A particle $A$ has charge $ + q$ and a particle $B$ has charge $ + \,4q$ with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference, the ratio of their speed $\frac{{{v_A}}}{{{v_B}}}$ will become

$2:1$

$1:2$

$1:4$

$4:1$

Solution

(b) Using $v = \sqrt {\frac{{2QV}}{m}} $ $==>$ $v \propto \sqrt Q $ $==>$ $\frac{{{v_A}}}{{{v_B}}} = \sqrt {\frac{{{Q_A}}}{{{Q_B}}}} = \sqrt {\frac{q}{{4q}}} = \frac{1}{2}$

Standard 12

Physics

A negatively charged plate has charge density of $2 \times {10^{ – 6}}\,C/{m^2}$. The initial distance of an electron which is moving toward plate, cannot strike the plate, if it is having energy of $200\,eV$

hard

$(a)$ Determine the electrostatic potential energy of a system consisting of two charges $7 \;\mu C$ and $-2\; \mu C$ (and with no external field) placed at $(-9 \;cm , 0,0)$ and $(9\; cm , 0,0)$ respectively.

$(b)$ How much work is required to separate the two charges infinitely away from each other?

$(c)$ Suppose that the same system of charges is now placed in an external electric field $E=A\left(1 / r^{2}\right) ; A=9 \times 10^{5} \;C m ^{-2} .$ What would the electrostatic energy of the configuration be?

medium

A positive point charge is released from rest at a distance $r_0$ from a positive line charge with uniform density. The speed $(v)$ of the point charge, as a function of instantaneous distance $r$ from line charge, is proportional to

hard

(JEE MAIN-2019)

Three point charges $Q, 4Q $ and $16Q $ are placed on a straight line $9$ $cm$ long. Charges are placed in such a way that the system has minimum potential energy. Then

hard

When a charge of $3\, coulomb$ is placed in a uniform electric field, it experiences a force of $3000\, Newton$. Within this field, potential difference between two points separated by a distance of $1\, cm$ is……..$volts$

easy

A particle $A$ has charge $ + q$ and a particle $B$ has charge $ + \,4q$ with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference, the ratio of their speed $\frac{{{v_A}}}{{{v_B}}}$ will become

Solution

Similar Questions

A negatively charged plate has charge density of $2 \times {10^{ – 6}}\,C/{m^2}$. The initial distance of an electron which is moving toward plate, cannot strike the plate, if it is having energy of $200\,eV$

A positive point charge is released from rest at a distance $r_0$ from a positive line charge with uniform density. The speed $(v)$ of the point charge, as a function of instantaneous distance $r$ from line charge, is proportional to

Three point charges $Q, 4Q $ and $16Q $ are placed on a straight line $9$ $cm$ long. Charges are placed in such a way that the system has minimum potential energy. Then

When a charge of $3\, coulomb$ is placed in a uniform electric field, it experiences a force of $3000\, Newton$. Within this field, potential difference between two points separated by a distance of $1\, cm$ is……..$volts$

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