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Electric potential Quiz 2

physics

 Anna found that an electron which is placed at an infinitesimal distance along the x-axis is having the momentum as $6\text{ }kgm{{s}^{-1}}$. What will be the value of electric potential? Let us assume that the velocity of light is $C=3\times {{10}^{8}}\text{ }m{{s}^{-1}}$.

  The product of momentum and velocity of light can be defined as the energy. The product of the charge and the electric potential can be defined as the electric potential energy. 

  Let $P$ be the momentum. $ \\ $ The equation of energy, $U=PC$ $ \\ $ The electric potential can be found by the equation, $U=qV$ $ \\ $ Comparing both the equations, $PC=qV$ $ \\ $ Substituting the values in it, $ \\ 6\times 3\times {{10}^{8}}=1.6\times {{10}^{-19}}\times V \\ \Rightarrow 18\times {{10}^{8}}=1.6\times {{10}^{-19}}\times V \\ $ Rearranging, $ \\ V=\dfrac{18\times {{10}^{8}}}{1.6\times {{10}^{-19}}} \\ \therefore V=11.25\times {{10}^{27}}\text{ }V$

 The work done to move a charge in an electric field of value mentioned as $4\text{ V}{{\text{m}}^{-1}}$, found to be $-6\text{ }J$. Calculate the angle at which the charge entered the electric field if the length of travel be $2\text{ }mm$. $($Let the value of charge be $3\times {{10}^{3}}\text{ }C)$

 The work done on a moving charge will be the negative of the product of the charge, electric field, length of the electric field region and the cosine of the angle. 

  The equation of energy, $ \\ W=-qEdx\cos \theta \\ $ Substituting the values, $ \\ -6=-3\times {{10}^{3}}\times 4\times 2\times {{10}^{-3}}\cos \theta \\ $ Rearranging the values, $ \\ \cos \theta =0.25 \\ \therefore \theta ={{\cos }^{-1}}\left( 0.25 \right) \\ $ Therefore the value of angle, $ \\ \theta =75.52{}^\circ $

  A particle sitting in a three dimensional space is having a value of electric $3x\hat{i}+6y\hat{j}+9z\hat{k}$. Find out the coordinates of the force on the particle if the charge is $2\text{ }C$.

 The negative of the change in potential with respect to distance will be the electric field. 

 The equation of electric field, $ \\ E=-\dfrac{dV}{dr} \\ \Rightarrow E=-\left( \dfrac{dV}{dx},\dfrac{dV}{dy},\dfrac{dV}{dz} \right) \\ $ The value of electric field, $ \\ E=-\left[ \dfrac{d}{dx}\left( 3x\hat{i}+6y\hat{j}+9z\hat{k} \right)+\dfrac{d}{dy}\left(3x\hat{i}+6y\hat{j}+9z\hat{k} \right)+\dfrac{d}{dz}\left( 3x\hat{i}+6y\hat{j}+9z\hat{k} \right) \right] \\ $ Taking the derivatives, $ \\ E=-\left( \dfrac{d}{dx}\left( 3x\hat{i}+6y\hat{j}+9z\hat{k} \right),\dfrac{d}{dy}\left( 3x\hat{i}+6y\hat{j}+9z\hat{k} \right),\dfrac{d}{dz}\left( 3x\hat{i}+6y\hat{j}+9z\hat{k} \right) \right) \\ \Rightarrow E=-\left( \dfrac{d}{dx}\left( 3x \right),\dfrac{d}{dy}\left( 6y \right),\dfrac{d}{dz}\left( 9z \right) \right) \\ \therefore E=-\left( 3,6,9 \right)\text{ }V{{m}^{-1}} \\ $ The equation of force, $F=qE$ $ \\ $ Substituting the values, $ \\ F=-2\times \left( 3,6,9 \right)\text{ }N \\ \therefore F=-\left( 6,12,18 \right)\text{ }N$

 Let the electric field due to an electron in a region be $4\hat{i}+6\hat{j}$. Calculate the change in potential energy at this location.

  The negative of the dot product of the electric field and the area element can be defined as the change in potential energy.

 The equation of electric potential, $dV=-E\cdot dr$ $ \\ $ The area element can be written as, $dr=dx+dy$ $ \\ $ Substituting this in the equation, $ \\ dV=-E\cdot \left( dx+dy \right) \\ $ Substituting the values in it, $ \\ dV=-\left[ \left( 4\hat{i}+6\hat{j} \right)\cdot \left( dx+dy \right) \right] \\ \Rightarrow dV=-\left[ 4+6 \right] \\ \therefore dV=-10\text{ }V \\ $ The change in potential energy can be found as, $dU=qdV$ $ \\ $ Substituting the values, $ \\ dU=1.6\times {{10}^{-19}}\times -10 \\ \therefore dU=-1.6\times {{10}^{-18}}\text{ }J$

 The electric field in a region has been found to be $4x+3y\text{ }V{{m}^{-1}}$. The body was travelling from $A=4$ to $B=9$ in a two dimensional space. Find out the mass of the body if the potential energy is equivalent to the work done by the electric field. Let us assume that the charge of the body is $2\text{ }C$ and the height of the body is $4\text{ }m$.

 The potential energy can be found by taking the product of mass of the body, acceleration due to gravity and the height at which the body is placed.

 The equation of work done can be written as, $ \\ W=-q\int\limits_{A}^{B}{E\cdot dr} \\ $ Substituting the values in it will give, $ \\ W=-2\int\limits_{4}^{9}{\left( 4x+3y \right)\cdot \left( dx+dy \right)} \\ \Rightarrow W=-2\int\limits_{4}^{9}{\left( 4xdx+3ydy \right)} \\ \Rightarrow W=-2\left[ 2{{x}^{2}}+\dfrac{3}{2}{{y}^{2}} \right]_{4}^{9} \\ \Rightarrow W=-2\left( \left[ 162+121.5 \right]-\left[ 32+24 \right] \right) \\ \Rightarrow W=-2\left( 283.5-56 \right) \\ \therefore W=227.5\text{ }J \\ $ If this is equivalent to potential energy of the body, $W=U$ $ \\ $ The equation of potential energy, $U=mgh$ $ \\ $ Substituting the values in it, $ \\ 227.5=m\times 10\times 4 \\ \Rightarrow m=\dfrac{227.5}{40} \\ \therefore m=5.6875\text{ }kg$