A charged particle of mass 5 × {10^{ - 5}},kg is held stationary in space by placing it in an elect

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1. Electric Charges and Fields

easy

A charged particle of mass $5 \times {10^{ - 5}}\,kg$ is held stationary in space by placing it in an electric field of strength ${10^7}\,N{C^{ - 1}}$ directed vertically downwards. The charge on the particle is

$ - 20 \times {10^{ - 5}}\,\mu C$

$ - 5 \times {10^{ - 5}}\,\mu C$

$5 \times {10^{ - 5}}\,\mu C$

$20 \times {10^{ - 5}}\,\mu C$

Solution

(b) $QE = mg$ $ \Rightarrow \,Q = \frac{{mg}}{E} = \frac{{5 \times {{10}^{ – 5}} \times 10}}{{{{10}^7}}}= 5 \times 10^{-5}\,\mu C.$
Since electric field is acting downward so for balance charge must be negative.

Standard 12

Physics

A positively charged thin metal ring of radius $R$ is fixed in the $xy – $ plane with its centre at the $O$. A negatively charged particle $P$ is released from rest at the point $(0,\,0,\,{z_0})$, where ${z_0} > 0$. Then the motion of $P$ is

hard

(IIT-1998)

The unit of intensity of electric field is

easy

Three charges are placed as shown in figure. The magnitude of $q_1$ is $2.00\, \mu C$, but its sign and the value of the charge $q_2$ are not known. Charge $q_3$ is $+4.00\, \mu C$, and the net force on $q_3$ is entirely in the negative $x-$ direction. The magnitude of $q_2$ is

medium

As shown in the figure, a particle A of mass $2\,m$ and carrying charge $q$ is connected by a light rigid rod of length $L$ to another particle $B$ of mass $m$ and carrying charge $-q.$ The system is placed in an electric field $\vec E$ . The electric force on a charge $q$ in an electric field $\vec E$ is $\vec F = q \vec E $ . After the system settles into equilibrium, one particle is given a small push in the transverse direction so that the rod makes a small angle $\theta_0$ with the electric field. Find maximum tension in the rod.

normal

For the given figure the direction of electric field at $A$ will be