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This is one of our many quizzes on Electric potential energy. Click on this to access more such competitive quizzes on Electric Potential and Capacitance.

Electric potential energy Quiz 2

physics

 A metal cube's hollow side with length $''2\,m''$ is placed at $x=5\,m$ from the origin. An electric field $E=2.5\hat{i}\,N/C$ exists in space. What will be the potential at the point $x=6\,m$ from the origin, given that potential at origin is $2V.$ 

 Because of electrostatic shielding, metal structures have a continuous potential all over their surface and inside.Also for a uniform electric field, $E=\dfrac{-\Delta V}{\Delta x}$. 

  $\text{The potential at }x=5m\text{ from where the block starts be }{{V}_{5m}}. \\ E=\dfrac{-\Delta V}{\Delta x} \\ \Rightarrow 2.5=\dfrac{-\left( {{V}_{5m}}-{{V}_{0}} \right)}{5-0} \\ \Rightarrow 2.5=\dfrac{-{{V}_{5m}}+2}{5} \\ \Rightarrow {{V}_{5m}}=-10.5V \\ \text{As, the electric field inside a metallic object is always zero} \\ \text{and the potential is constant due to electrostatic shielding,} \\ \text{the potential at }x=6m,{{V}_{6m}} \\ \therefore {{V}_{6m}}={{V}_{5m}}=-10.5\,V$

If direction of an electric field in some space is in positive $X$-direction and a charge of $10\,\mu C$ is travelling along the $x-axis$ then its electric potential energy will

 Electric potential decreases in the direction of the electric field as $\text{V}=-\int{\vec{E}.d\vec{r}}$. Also, electric potential energy $U=qV$. 

  $\text{As V}=-\int{\vec{E}.d\vec{r}} \\ \text{And U=qV} \\ \therefore \text{ U=}-q\int{\vec{E}.d\vec{r}} \\ \text{Thus a negative sign shows that the Electric potential energy } \\ \text{decreases for a charge moving in the direction of Electric field}\text{.}$

For a given electric field, the locus of points having equal potential will  

 Equipotential surface is the locus of points of equal potential in an electric field. Work done on moving a charge anywhere on an equipotential surface is zero. 

 &nbsp;$\text{The locus of points having equal potential is the equipotential surface in an electric field.} \\ \text{As work done on moving a charge anywhere on an equipotential surface is zero,} \\ \text{So }{{\text{W}}_{electric}}=0 \\ {{{\vec{F}}}_{electric}}.\vec{r}=0 \\ \text{As neither }\left| {{{\vec{F}}}_{electric}} \right|\text{ and }\left| {\vec{r}} \right|\text{is zero then} \\ \cos \theta =0 \\ \text{Thus an equipotential and surface is always perpendicular the electric field lines}\text{.}$

 Two semi discs of surface charge density $\sigma $ are placed $'R'$ distance apart from a point as shown below. The potential at this point $O$ is<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7d83a321-d3a1-4a6e-9cff-4d68ed1a1b23173206595976677816.png"/> 

 The potential due to a circular disc on its axis is $V=\dfrac{2\sigma }{4\pi {{\varepsilon }_{0}}}\left( \sqrt{{{R}^{2}}+{{x}^{2}}}-\sqrt{{{x}^{2}}} \right)$. Potential due to a semi circular disc its axis would be $=\dfrac{V}{2}=\dfrac{\sigma }{4\pi {{\varepsilon }_{0}}}\left( \sqrt{{{R}^{2}}+{{x}^{2}}}-\sqrt{{{x}^{2}}} \right)$. 

 $\text{Potential at }O\text{ due semi disc at }A\text{ is} \\ \text{V}_{o}^{'}=\dfrac{\sigma }{4\pi {{\varepsilon }_{0}}}\left( \sqrt{{{R}^{2}}+{{R}^{2}}}-\sqrt{{{R}^{2}}} \right) \\ \text{V}_{o}^{'}=\dfrac{\sigma }{4\pi {{\varepsilon }_{0}}}\left( R\sqrt{2}-R \right) \\ \text{V}_{o}^{'}=\dfrac{\sigma R\left( \sqrt{2}-1 \right)}{4\pi {{\varepsilon }_{0}}} \\ \text{Similarly potential at }O\text{ due to semi disc at }B\text{ will also be} \\ \text{V}_{o}^{''}=\dfrac{\sigma R\left( \sqrt{2}-1 \right)}{4\pi {{\varepsilon }_{0}}} \\ \text{Total potential at O} \\ \text{V}=V_{o}^{'}+V_{o}^{''}=\dfrac{\sigma R\left( \sqrt{2}-1 \right)}{4\pi {{\varepsilon }_{0}}}+\dfrac{\sigma R\left( \sqrt{2}-1 \right)}{4\pi {{\varepsilon }_{0}}} \\ {{V}_{o}}=\dfrac{2\sigma R}{4\pi {{\varepsilon }_{0}}}\left( \sqrt{2}-1 \right)$

A charge of magnitude $+2q$ is displaced from position $O$ to $P$ as shown in figure.In this process magnitude of electric potential energy will<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/16a0aeb0-57a1-41cc-b06f-80646bff1fd61589400565588148596.png"/> 

 Potential energy between two points for moving a charge $Q$ is $Q\Delta V$, when $\Delta V$ is the difference of potential of the two points.  

 $\text{Charge at }A\text{ and }B\text{ are equidistant from point O and P due to symmetry.} \\ \text{Thus potential due to them will also be equal.} \\ \text{The same happens due to charges D and C.} \\ \text{From pythagoras theorem}\text{.} \\ A{{O}^{2}}=A{{M}^{2}}+O{{M}^{2}} \\ \Rightarrow AO=\sqrt{\dfrac{{{R}^{2}}}{{{\left( 2 \right)}^{2}}}+\dfrac{{{R}^{2}}}{{{\left( 2 \right)}^{2}}}}=\dfrac{R}{\sqrt{2}} \\ \Rightarrow AO=OB=AP=BP\text{ due to symmetry} \\ \text{Thus }{{\text{V}}_{o}}\text{ due to }A\text{ and }B\text{ is } \\ {{\text{V}}_{o}}=\dfrac{kq}{\dfrac{R}{\sqrt{2}}}+\dfrac{kq}{\dfrac{R}{\sqrt{2}}}=\dfrac{2\sqrt{2}kq}{R} \\ \text{Potential at P due to charge A and B} \\ {{\text{V}}_{p}}=\dfrac{kq}{\dfrac{R}{\sqrt{2}}}+\dfrac{kq}{\dfrac{R}{\sqrt{2}}}=\dfrac{2\sqrt{2}kq}{R} \\ \text{As }{{V}_{p}}-{{V}_{o}}=0 \\ \text{Because of A and B } \\ \text{Thus } \\ \Delta \text{U=2q}\left( {{V}_{p}}-{{V}_{o}} \right)=0 \\ \text{By pythagoras theorem} \\ \Rightarrow O{{D}^{2}}=D{{N}^{2}}+O{{N}^{2}} \\ \Rightarrow OD=\sqrt{D{{N}^{2}}+O{{N}^{2}}} \\ \Rightarrow OD=\sqrt{{{\left( \dfrac{R}{2} \right)}^{2}}+{{\left( \dfrac{R}{2} \right)}^{2}}} \\ \Rightarrow OD=\dfrac{R}{\sqrt{2}} \\ \text{And }D{{P}^{2}}=P{{N}^{2}}+D{{N}^{2}} \\ DP=\sqrt{{{\left( \dfrac{3R}{2} \right)}^{2}}+{{\left( \dfrac{R}{2} \right)}^{2}}} \\ \Rightarrow DP=\dfrac{R\sqrt{10}}{2} \\ \text{Due to symmetry }DP=CP=\dfrac{R\sqrt{10}}{2} \\ \text{And }OD=OC=\dfrac{R}{\sqrt{2}} \\ \text{Potential at O due to C and D} \\ \text{V}_{0}^{1}=\dfrac{kq}{\dfrac{R}{\sqrt{2}}}+\dfrac{kq}{\dfrac{R}{\sqrt{2}}}=\dfrac{2\sqrt{2}q}{R} \\ \text{Potential at P due to C and D} \\ \text{V}_{p}^{1}=\dfrac{kq}{\dfrac{R\sqrt{10}}{2}}+\dfrac{kq}{\dfrac{R\sqrt{10}}{2}}=\dfrac{4kq}{R\sqrt{10}} \\ \text{Charge in potential from O to P due to C and D}=V_{p}^{1}-V_{0}^{1} \\ \Rightarrow \dfrac{4kq}{R\sqrt{10}}-\dfrac{2\sqrt{2}kq}{R} \\ \Rightarrow \dfrac{2kq}{R}\left( \dfrac{2}{\sqrt{10}}-\sqrt{2} \right) \\ \text{Potential energy}=2q\Delta v \\ \Rightarrow \text{Potential energy}=2q\left[ \dfrac{2kq}{R}\left( \dfrac{2}{\sqrt{10}}-\sqrt{2} \right) \right] \\ \therefore \text{Potential energy}=\dfrac{4kq}{R}\left( \dfrac{2}{\sqrt{10}}-\sqrt{2} \right) \\ \text{This is the total potential energy}$