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This is one of our many quizzes on Electric field. Click on this to access more such competitive quizzes on Electrostatics and Magnetism.

Electric field Quiz 2

physics

Value of electric field at point $''O''$ will be $($where, $P=4q)$ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7fe01a7e-f6d4-4cab-8c22-bb392b0ab9546535804360854627309.png"/> 

 Use the formula of Electric field at a point in a plane of an electric dipole then value of $E=\dfrac{KP}{{{r}^{3}}}\sqrt{1+3{{\cos }^{2}}\theta }$ where$K=\dfrac{1}{4\pi {{\varepsilon }_{0}}}$.

  Electric field due to a dipole at any point on it’s plane is $ \\ {{E}_{0}}=\dfrac{KP}{{{r}^{3}}}\sqrt{3{{\cos }^{2}}\theta +1} \\ $ Also $\theta =30{}^\circ,$  $ \\ \Rightarrow {{E}_{0}}=\dfrac{P}{4\pi {{\varepsilon }_{0}}{{L}^{3}}}\sqrt{3\times {{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}+1} \\ \Rightarrow {{E}_{0}}= \dfrac{\sqrt{13}}{2}\times \dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{P}{{{L}^{3}}} \\ $ As dipole moment $P$ is $P=4q$ $ \\ \Rightarrow {{E}_{0}}= \dfrac{\sqrt{13}}{2}\times \dfrac{1}{4\pi {{\varepsilon }_{0}}}\times \dfrac{4q}{{{L}^{3}}} \\ \therefore {{E}_{0}}= \dfrac{q\sqrt{13}}{2\pi {{\varepsilon }_{0}}{{L}^{3}}}$

An electron is placed at a distance of $1$ meter from a line of infinite length and $\lambda $ charge per unit length. The magnitude of force experience by electron is $ \\ $  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9f3426e0-902f-4bff-941e-93aaec835bdf5458565040111659850.png"/>

 Magnitude of electric field at a distance $''r''$ from an infinite charged line is $\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}r}$. 

 $\lambda =\text{Charge density of line} \\ r=\text{Distance from wire} \\ E=\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}r} \\ \Rightarrow F=qE=-eE \\ \therefore \left| F \right|=eE=\left( \dfrac{\lambda e}{2\pi {{\varepsilon }_{0}}} \right)$

 The angle with which the vertical of an electric field needed to provide a horizontal acceleration of $2\,m{{s}^{-2}}$ for a unit charge of unit mass in air on earth would be- 

 Force on a charge $Q$ due to an electric field $E$ is $F=qE$. Draw the $FBD$ and solve force equations along the $x$ and $y$ axis separately. 

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3cf25fdc-69ee-404d-a838-eb67d08256ff4419206872108472717.png"/>  $ \\ $ Let $\theta$ be the angle of the electric field with the vertical. From the F.B.D  $ \\ \text{ }{{F}_{e}}\cos \theta =mg\text{     }...\left( 1 \right) \\ \text{But, }{{F}_{e}}=qE,\text{ }q=1C \\ \Rightarrow {{F}_{e}}=E \\ $ Equation $\left( 1 \right)$ becomes, $ \\ E\cos \theta =\left( 1 \right)\left( 10 \right) \\ \Rightarrow E\cos \theta =10\text{     }...\left( 2 \right) \\ $ And from the F.B.D working along the horizontal direction, $ \\ \Rightarrow {{F}_{e}}\sin \theta =\max  \\ \Rightarrow  E\sin \theta =\left( 1 \right)\left( 2 \right) \\ \Rightarrow E\sin \theta =2\text{     }...\left( 3 \right) \\ $  Dividing $\left( 3 \right)$ by $\left( 2 \right),$ $ \\ \Rightarrow \tan \theta =\dfrac{2}{10}=\dfrac{1}{5} \\ \therefore \theta ={{\tan }^{-1}}\left( \dfrac{1}{5} \right)$

Two infinite charged sheets with surface charge density $\left( \sigma  \right)$ are separated by distance $\left( L \right)$ as shown in figure. One sheet is positively charged while the other is negatively charged. If measured from the negative plate, the Electric field would be zero at a distance of $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/052f2c7e-40e8-45c4-8f20-68cc8b0d9a405368501152696915109.png"/> 

 Electric field at distance $''r''$ from infinite charged sheet is $\dfrac{\sigma }{2{{\varepsilon }_{0}}}$, where $''\sigma ''$ is charge density of sheet. So the Electric field is uniform or it is independent of the distance from the sheets. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ccbfc724-95c6-4a0e-950f-16e27b37789d2480670393791616156.png"/> <br>$ \\  $ Due to negative charge sheet value of $E$ at distance $r$ is $\dfrac{-\sigma }{2{{\varepsilon }_{0}}}$, due to positive charge sheet $ \\ $  Also, the electric field due to an infinite charged plane sheet is independent of the distance from the sheet $ \\ E=\dfrac{\sigma }{2{{\varepsilon }_{0}}} \\ $ The electric field in the region between the sheets is in the same direction due to both plates. Thus, $ \\ {{E}_{net}}={{E}_{1}}+{{E}_{2}} \\ \Rightarrow E=\dfrac{-\sigma }{2{{\varepsilon }_{0}}}+\dfrac{\sigma }{2{{\varepsilon }_{0}}}=0\\ $ As the Electric field is independent of distance, i.e it is uniform $ \\ $ The Electric field is always opposite in the exterior region of the two plates. $ \\ $ Thus for any distance $x=0$ to $x=L$ from the negatively charged plate, the net Electric field will be zero.

A given hollow dielectric shell of inner and outer radius ${{r}_{1}}$ and ${{r}_{2}}$ is given a charge of volume charge density $\rho \text{ }c{{m}^{-2}}$. Later, a conducting thin shell of radius $R > {{r}_{1}},{{r}_{2}}$ with surface charge density $\sigma $ is placed concentrically around the dielectric sphere. The electric field between the region of space of the conducting shell and dielectric at a point of distance $'d'$ from the circle is  

 Electric field due to a charged conductor inside itself is always zero, while a dielectric sphere behaves like a point charge for a point outside its surface and the electric field of a point charge is $E=\dfrac{kQ}{{{r}^{2}}}$.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/01b9068a-bdc3-4b34-9ecb-8972d809db4d7862556023351450450.png"/> $ \\ $ By the law of superposition, total electric field $E$ at point $D$ is $ \\ $ ${{{\vec{E}}}_{net}}={{{\vec{E}}}_{dielectric}}+{{{\vec{E}}}_{conductor}} \\ {{{\vec{E}}}_{dielectric}}= $ Electric field of dielectric shell $ \\ $ ${{{\vec{E}}}_{conductor}} = $ Electric field of conducting shell $ \\ $ But ${{E}_{conductor}}=0$ $ \\ $ As the dielectric behaves like a point charge for electric field outside its surface. $ \\ {{Q}_{T}}=\rho \left( {{V}_{outer}}-{{V}_{inner}} \right) \\ {{Q}_{T}}=\rho \left( \dfrac{4}{3}\pi {{r}_{2}}^{3}-\dfrac{4}{3}\pi {{r}_{1}}^{3} \right) \\ \therefore {{Q}_{T}}=\dfrac{4}{3}\pi \rho \left( {{r}_{2}}^{3}={{r}_{1}}^{3} \right) \\ \Rightarrow {{E}_{net}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{Q}_{T}}}{{{d}^{2}}}+0 \\ \Rightarrow {{E}_{net}}=\dfrac{4}{4\pi {{\varepsilon }_{0}}}\dfrac{\dfrac{4}{3}\pi \rho \left( {{r}_{2}}^{3}-{{r}_{1}}^{3} \right)}{{{d}^{2}}} \\ \therefore {{E}_{net}}=\dfrac{\rho }{3{{\varepsilon }_{0}}}\dfrac{\left( {{r}_{2}}^{3}-{{r}_{1}}^{3} \right)}{{{d}^{2}}}$