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This is the second of our many quizzes on Electric charge. Click on this to access more such competitive quizzes on Electrostatics and Magnetism.

Electric charge Quiz 2

physics

Find out the value of charge $q$ from the following figure if the electric field due to the sphere be $6\text{ }V{{m}^{-1}}$?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a416f0b9-ab6c-4433-9b9b-c9c814aae1766329907659305907694.png"/> 

The electric field will be the ratio of the charge to the permittivity. 

 $\text{The electric field can be found as,}\\E=\dfrac{{{q}'}}{{{\varepsilon }_{0}}}\\\text{The net charge in the sphere will be,}\\{q}'=+5q+q-3q=+3q\\\text{Therefore,}\\6=\dfrac{3{q}'}{{{\varepsilon }_{0}}} \\\therefore {q}'=2{{\varepsilon }_{0}}\text{ }C$ 

When a charged particle is revolving around a nucleus, the radius of the orbit has been given as $4\times {{10}^{-15}}\text{ }m$. The velocity at which the particle rotates is given as $2.8\times {{10}^{-6}}\text{ }m{{s}^{-1}}$. The mass of the particle is $3\times {{10}^{-19}}\text{ kg}$ and the electric field produced here will be $5\text{ }V{{m}^{-1}}$. Calculate the charge of the particle. 

The centripetal force will be equal to the electrostatic force. 

  $\text{The centripetal force will be equal to the electrostatic force,}\\\dfrac{m{{v}^{2}}}{r}=qE \\\Rightarrow q=\dfrac{m{{v}^{2}}}{Er} \\\Rightarrow q=\dfrac{3\times {{10}^{-19}}\times {{\left( 2.8\times {{10}^{-6}} \right)}^{2}}}{5\times 4\times {{10}^{-15}}} \\\therefore q=1.176\times {{10}^{-16}}\text{ }C$

 The surface charge density of a slab of area $3\text{ }{{m}^{2}}$ has been given as $5\times {{10}^{6}}\text{ }C{{m}^{-2}}$. Sumith wants to know the amount of charge included in it. What will be the conclusion at which he reaches?

 The Ratio of the charge and the area can be defined as the surface charge density. 

$\text{Surface charge density,}\\\sigma =\dfrac{q}{A} \\\Rightarrow 5\times {{10}^{6}}=\dfrac{q}{3} \\\therefore q=15\times {{10}^{6}}\text{ }C$ 

The electric charges are having a major application in photocopiers and printers. What will be the reason for this?

 The static electric charges will be created there.  

 The toner $(\text{ dry black powder})$ will get negatively charged at first and this negative charge gets attracted to the positive image made on the drum of the printers from which it gets pulled out by a highly charged paper. Therefore the static electric charges will attract the toner to the paper.

The voltage across a resistor having resistance $5\text{ }\Omega $ has been mentioned as $4\text{ }V$. What will be the amount of charge flowing for a time interval of $t=15\text{ s}$? 

The voltage will be the product of multiplying the current by the resistance, and the current will be the rate of charge movement.

$\text{The voltage will be,}\\V=IR \\\Rightarrow I=\dfrac{V}{R} \\\text{The current can be found by the equation,}\\I=\dfrac{Q}{t}\\\text{Comparing both of these,}\\V=IR \\\Rightarrow \dfrac{V}{R}=\dfrac{Q}{t} \\\Rightarrow Q=\dfrac{Vt}{R} \\\text{Substituting the values,}\\Q=\dfrac{4\times 15}{5} \\\therefore Q=12\text{ }C$