Q: If \${ 10 }^{ 9 }\$ electron move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?

Given,\${ 10 }^{ 9 }\$ electron takes \$1 sec\$So, \$1\$ electron takes \$\rightarrow \cfrac { 1 }{{ 10 }^{ 9 }}sec\$And we know, \$1 C= 6.242 \times { 10 }^{ 18 } electron\$\$\therefore 6.242 \times { 10 }^{ 18 }electrons\$ takes \$\rightarrow \cfrac{6.242 \times { 10 }^{ 18 }}{{ 10 }^{ 9 }}\$\$= 6.242 \times { 10 }^{ 9 }sec \$\$= \cfrac{6.242 \times { 10 }^{ 9 }}{60 \times 60 \times 24 \times 365}years\$\$=197.93 years\$\$ \simeq 198years\$If \${ 10 }^{ 9 }\$ electrons move out of a body to another body every second, then \$198 years\$ is required to get a total charge of \$1C\$ on the other body.

Q: Derive the formula for the electric potential energy of an electric dipole in a uniform electric field.

Potential energy of dipole in uniform electric field.\$\vec {\tau}=\vec p\times \vec E\$work done in rotating dipole by small angle\$dw=\tau d\theta\$\$=-p\varepsilon \sin \theta d\theta\$The change in potential energy \$dU=-dW=pE\sin \theta d\theta\$If \$\theta \rightarrow 90^o\$ to \$\theta^o\$\$U(\theta) -U(90^o)=\displaystyle \int_{90^o}^{0^o}{P\varepsilon \sin \theta d\theta}\$\$=P\varepsilon [-\cos \theta]_{90}^{0}\$\$=-\vec P.\vec E\$So\$\boxed{U(\theta)=-\vec P. \vec E}\$

Q: A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of (i) capacitance.(ii) potential difference between the plates.(iii) electric field between the plates.(iv) the energy stored in the capacitor.

\$(i)\$ The capacitance increases as the dielectric constant \$K > 1\$. \$(ii)\$ Potential difference \$V =\dfrac{Q}{C}\$. As \$C\$ increases and \$Q\$ remains the same since the battery is disconnected, the p.d. between the plates decreases.\$(iii)\$ Electric field \$E =\dfrac{V}{d}\$ where \$V\$ is the p.d. and \$d\$ the separation between the plates. As \$V\$ decreases and d remains the same, electric field also decreases.\$(iv)\$ Energy stored in a capacitor \$U =\dfrac{1}{2}\dfrac{{Q}^{2}}{C}\$. As \$Q\$ is constant and \$C\$ increases, \$U\$ decreases.

Q: Derive the relation between electric field and electric potential due a point charge.

Consider two points \$A\$ and \$B\$ separated by a small distance \$dx\$ in an electric field.Since \$dx\$ is small, the electric field \$E\$ is assumed to be uniform along \$AB\$. The force acting on a unit positive charge at A is equal to \$E\$.Now, the work done in moving a unit positive charge from \$A\$ to \$B\$ against the electric field is \$dW=-E\times dx\$.The negative sign shows that the work is done against the direction of the field. Since the work done is equal to potential difference \$dV\$ between \$A\$ and \$B\$ then \$dV=-E\times dx\$ or \$E=-\dfrac{dV}{dx}\$Thus, the electric field at a point is the negative of the potential gradient at that point.

Q: (a) Deduce the expression for the potential energy of a system of two charges \$q_1\$ and \$q_2\$ located \$\overrightarrow{r_1}\$ and \$\overrightarrow{r_2}\$, respectively, in an external electric field.(b) Three point charges, + Q + 2Q and - 3Q are placed at the vertices of an equilateral triangle ABC of side l. If these charges are displaced to the mid-point \$A_1, B_1\$ and \$C_1\$, respectively, find the amount of the work done in shifting the charges to the new locations.

\$\left( a \right) \rightarrow (i)\$ work done to bring a charge q in the electric field at a distance \${ r }_{ 1 }={ q }_{ 1 }V\left( { r }_{ 1 } \right) \$ \$(ii)\$ work done to bring a charge q in the electric field at a distance \${ r }_{ 2 }={ q }_{ 2 }V\left( { r }_{ 2 } \right) \$ \$(iii)\$ work done on \${ q }_{ 2 }\$ to move it against the force of \${ q }_{ 1 }=\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } \$\$\therefore \$ The potential energy of the system \$={ q }_{ 1 }V\left( { r }_{ 1 } \right) +{ q }_{ 2 }\left( V \right) \left( { r }_{ 2 } \right) +\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } \$\$(b)\$ work done in moving charges \$=\$ Potential of the initial position - Potential of the final position.Now, Potential of the initial system \$={ V }_{ 12 }+{ V }_{ 23 }+{ V }_{ 31 }={ 2KQ }^{ 2 }+\left( { -3KQ }^{ 2 } \right) +\left( -6K{ Q }^{ 2 } \right) =-7K{ Q }^{ 2 }J\$potential energy after moving to position, \${ A }_{ 1 },{ B }_{ 1 }\& { C }_{ 1 }\$\$=\dfrac { K\left( 2Q \right) \left( Q \right) }{ 1/2 } +\dfrac { K\left( Q \right) \left( -3Q \right) }{ 1/2 } +\dfrac { K\left( -3Q \right) \left( 2Q \right) }{ 1/2 } \$\$=4{ KQ }^{ 2 }-6{ KQ }^{ 2 }-12{ KQ }^{ 2 }=-14K{ Q }^{ 2 }J\$\$\therefore \$ work done \$=-7{ Q }^{ 2 }K-\left( -14K{ Q }^{ 2 } \right) =7K{ Q }^{ 2 }J\$

Question 1

Derive the expression for capacitance of parallel plate capacitor.

Accepted Answer

Let the two plates are kept parallel to each other separated be a distance $d$ and cross-sectional area of each plate is $A$.  Electric field by a single thin plate $E' = \dfrac{\sigma}{2\epsilon_o}$Total electric field between the plates $E = \dfrac{\sigma}{2\epsilon_o} +\dfrac{\sigma}{2\epsilon_o}$Or $E = \dfrac{\sigma}{\epsilon_o}$Or $E = \dfrac{Q}{A\epsilon_o}$Potential difference between the plates $V = Ed$$\implies$ $V = \dfrac{Qd}{A\epsilon_o}$Capacitance  $C = \dfrac{Q}{V}$Thus we get capacitance of parallel plate capacitor  $C  =\dfrac{A\epsilon_o}{d}$

Question 2

Two point charges $q_A = 3\mu C$ and $q_B =- 3 \mu C$ are located $20\ cm$ apart in vacuum(a) What is the electric field at the midpoint O of the line $AB$ joining the two charges?(b) If a negative test charge of magnitude $1.5\times10^{-9}C$ is placed at this point, what is the force experienced by the test charge?

Accepted Answer

(a) The situation is represented in the given figure. O is the mid-point of line AB.Distance between the two charges, $AB = 20 cm$$\therefore AO = OB = 10 cm$Net electric field at point $O = E$Electric field at point O caused by +3 C charge,$E_1 = \displaystyle \frac{3\times 10^{-6}}{4\pi\in_0(AO)^2}=\frac{3\times 10^{-6}}{4\pi\in_0(10\times10^{-2})^2}N/C$ along OBWhere,$\in_0=$ Permittivity of free space$\displaystyle \frac{1}{4\pi\in_0}=9\times 10^9Nm^2C^{-2}$Magnitude of electric field at point O caused by - 3 C charge,$E_2=\displaystyle \left|\frac{-3\times 10^{-6}}{4\pi\in_0(OB)^2}\right|=\frac{3\times 10^{-6}}{4\pi\in_0(10\times 10^{-2})^2}N/C$ along OB$\therefore E=E_1+E_2$$=\displaystyle 2\times \left[(9\times 10^9)\times \frac{3\times 10^{-6}}{(10\times 10^{-2})^2}\right]$ [As $E_1$ = $E_2$, the value is multiplied with 2]$=5.4 \times 10^6 $ N/ C   along OBTherefore, the electric field at mid-point O is $5.4 \times 10^6 $ N/ C  along OB.(b) A test charge of amount $q =1.5 \times 10^{-9} C$ is placed at mid-point O.$q =1.5 \times 10^{-9} C$Force experienced by the test charge $= F$$\therefore F = qE$$=1.5 \times 10^{-9} \times 5.4 \times 10^{6} $$=8.1 \times 10^{-3} N$The force is along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.Hence, the force experienced by the test charge is $8.1 \times 10^{-3} \: N$along OA.

Question 3

If ${ 10 }^{ 9 }$ electron move out of a body to another body every second, how much time is required to get a total charge of 1 C on the other body?

Accepted Answer

Given,${ 10 }^{ 9 }$ electron takes $1 sec$So, $1$ electron takes $\rightarrow \cfrac { 1 }{{ 10 }^{ 9 }}sec$And we know, $1 C= 6.242 \times { 10 }^{ 18 } electron$$\therefore 6.242 \times { 10 }^{ 18 }electrons$ takes $\rightarrow \cfrac{6.242 \times { 10 }^{ 18 }}{{ 10 }^{ 9 }}$$= 6.242 \times { 10 }^{ 9 }sec $$= \cfrac{6.242 \times { 10 }^{ 9 }}{60 \times 60 \times 24 \times 365}years$$=197.93 years$$ \simeq 198years$If ${ 10 }^{ 9 }$ electrons move out of a body to another body every second, then $198 years$ is required to get a total charge of $1C$  on the other body.

Question 4

Derive an expression for energy stored in a capacitor. In which form energy is stored?

Accepted Answer

Let us consider a capacitor of capacitance C and potential difference V between the plates.Let the charge on one plate be +q and -q on the other.Suppose the capacitor is being charged gradually.Now,at any stage the charge on capacitor is q.Therefore, the potential difference = $ \frac{q}{C} $Small amount of work doe in giving n additional charge dq to the capacitor isdW= $\dfrac{q}{C}*dq$Total work done in giving a charge Q to the capacitor  isW=$\int dW$W= $ \int_{0}^{Q} \frac{Q}{C}dq $ W = $\dfrac{Q^{2}}{C}$Energy = EE= $\dfrac{Q^2}{2C}$ = $\dfrac {CV^2}{2}$ = $\dfrac{QV}{2}$The energy is stored in the form of potential energy.

Question 5

Derive the formula for the electric potential energy of an electric dipole in a uniform electric field.

Accepted Answer

Potential energy of dipole in uniform electric field.$\vec {\tau}=\vec p\times \vec E$work done in rotating dipole by small angle$dw=\tau d\theta$$=-p\varepsilon \sin \theta d\theta$The change in potential energy $dU=-dW=pE\sin \theta d\theta$If $\theta \rightarrow 90^o$ to $\theta^o$$U(\theta) -U(90^o)=\displaystyle \int_{90^o}^{0^o}{P\varepsilon \sin \theta d\theta}$$=P\varepsilon [-\cos \theta]_{90}^{0}$$=-\vec P.\vec E$So$\boxed{U(\theta)=-\vec P. \vec E}$

Question 6

A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of (i) capacitance.(ii) potential difference between the plates.(iii) electric field between the plates.(iv) the energy stored in the capacitor.

Accepted Answer

$(i)$ The capacitance increases as the dielectric constant $K > 1$. $(ii)$ Potential difference $V =\dfrac{Q}{C}$.  As $C$ increases and $Q$ remains the same since the battery is disconnected, the p.d. between the plates decreases.$(iii)$ Electric field $E =\dfrac{V}{d}$ where $V$ is the p.d. and $d$ the separation between the plates. As $V$ decreases and d remains the same, electric field also decreases.$(iv)$ Energy stored in a capacitor $U =\dfrac{1}{2}\dfrac{{Q}^{2}}{C}$. As $Q$ is constant and $C$ increases, $U$ decreases.

Question 7

Derive an expression for the electric potential due to a point charge :

Accepted Answer

Consider the electric potential due to a point charge q, As we move from point A, at distance rA from the charge q, to point B, at distance rB from the charge q, the change in electric potential is$\Delta V_{BA}= V_B-V_A=- \int_A^B E. ds$$E.ds=[k\dfrac{q}{r^2}]\hat{r}.ds$$\hat{r}.ds=dr$Only the radial distance r determines the work done or the potential. We can move through any angle we like and, as long as the radial distance remains constant, no work is done or there is no change in the electric potential.$\Delta V_{BA}= V_B-V_A = - \int_{r_A}^{r_B}Edr=-\int_{r_A}^{r_B}[k\dfrac{q}{r^2}dr]$$\Delta V_{BA}= V_B-V_A=-kq[(-1)r^{-1}]_{r_A} ^{r_B}$$\Delta V_{BA}= V_B-V_A= kq [\dfrac{1}{r_B}-\dfrac{1}{r_B}]$This is the change in electric potential due to a point charge as we move from rA to rB.We could ask about the change in electric potential energy as we move a charge q' from radius rA to rB due to a point charge q$\Delta U_{BA}= kq'q[\dfrac{1}{r_B}-\dfrac{1}{r_A}]$As with gravitational potential energy, it is more convenient -- and, therefore, useful -- to talk about the electric potential energy or the electric potential relative to some reference point. We will choose that reference point to be infinity. That is,$r_A=\infty$That means we can then write the electric potential at some radius r as $V=kq\dfrac{1}{r}$

Question 8

Two charges $ 5 \times 10^{-8}\ C$ and $-3\times 10^{-8}\ C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Accepted Answer

Let p be the point of zero potential be p at distance r from the charge 1..$d=16\ cm$Electric potential, $v = q_1/4\pi\epsilon_or+ q_2/4\pi\epsilon_o(d-r)$for $v=0$, $\displaystyle \frac{q_1}{r}=-\frac{q_2}{d-r}$$\Rightarrow \displaystyle \frac{5\times 10^{-8}}{r}=\frac{-3\times 10^{-8}}{0.16-r}$$r=10\ cm$If, point p is outside,$v = q_1/4\pi\epsilon_or+ q_2/4\pi\epsilon_o(r-d)$so here, $r=40\ cm$

Question 9

Derive the relation between electric field and electric potential due a point charge.

Accepted Answer

Consider two points $A$ and $B$ separated by a small distance $dx$ in an electric field.Since $dx$ is small, the electric field $E$ is assumed to be uniform along $AB$. The force acting on a unit positive charge at A is equal to $E$.Now, the work done in moving a unit positive charge from $A$ to $B$ against the electric field is $dW=-E\times dx$.The negative sign shows that the work is done against the direction of the field. Since the work done is equal to potential difference $dV$ between $A$ and $B$ then $dV=-E\times dx$ or $E=-\dfrac{dV}{dx}$Thus, the electric field at a point is the negative of the potential gradient at that point.

Question 10

(a) Deduce the expression for the potential energy of a system of two charges $q_1$ and $q_2$ located $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$, respectively, in an external electric field.(b) Three point charges, + Q + 2Q and - 3Q are placed at the vertices of an equilateral triangle ABC of side l. If these charges are displaced to the mid-point $A_1, B_1$ and $C_1$, respectively, find the amount of the work done in shifting the charges to the new locations.

Accepted Answer

$\left( a \right) \rightarrow (i)$ work done to bring a charge q in the electric field at a distance ${ r }_{ 1 }={ q }_{ 1 }V\left( { r }_{ 1 } \right) $            $(ii)$ work done to bring a charge q in the electric field at a distance ${ r }_{ 2 }={ q }_{ 2 }V\left( { r }_{ 2 } \right) $            $(iii)$ work done on ${ q }_{ 2 }$ to move it against the force of ${ q }_{ 1 }=\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } $$\therefore $ The potential energy of the system $={ q }_{ 1 }V\left( { r }_{ 1 } \right) +{ q }_{ 2 }\left( V \right) \left( { r }_{ 2 } \right) +\dfrac { K{ q }_{ 1 }{ q }_{ 2 } }{ { r }_{ 12 } } $$(b)$ work done in moving charges $=$ Potential of the initial position - Potential of the final position.Now, Potential of the initial system $={ V }_{ 12 }+{ V }_{ 23 }+{ V }_{ 31 }={ 2KQ }^{ 2 }+\left( { -3KQ }^{ 2 } \right) +\left( -6K{ Q }^{ 2 } \right) =-7K{ Q }^{ 2 }J$potential energy after moving to position, ${ A }_{ 1 },{ B }_{ 1 }\& { C }_{ 1 }$$=\dfrac { K\left( 2Q \right) \left( Q \right)  }{ 1/2 } +\dfrac { K\left( Q \right) \left( -3Q \right)  }{ 1/2 } +\dfrac { K\left( -3Q \right) \left( 2Q \right)  }{ 1/2 } $$=4{ KQ }^{ 2 }-6{ KQ }^{ 2 }-12{ KQ }^{ 2 }=-14K{ Q }^{ 2 }J$$\therefore $ work done $=-7{ Q }^{ 2 }K-\left( -14K{ Q }^{ 2 } \right) =7K{ Q }^{ 2 }J$