Q: A charge of 2\ C is moved from a point 2\ m away from a charge of 1\ C to a point 1\ m away from that charge. The work done is:

Voltage due to 1\ C at a distance r is \dfrac {k \times 1}{r} .Total change in potential in moving 2\ C charge from 2\ m away to 1\ m away: V_2-V_1=\dfrac {k \times 1}{1}-\dfrac {k \times 1}{2} or, V_2-V_1=\dfrac{k \times 1}{2} = \Delta V So, work done =\Delta V Q =\dfrac {k \times 1}{2}\times 2=9\times 10^9\ J

Question 1

An infinite number of charges each equal to  'q'  are placed along the X-axis at  x = 1 ,  x = 2 ,  x = 4 ,  x = 8  The potential at the point  x = 0  due to this set of charges is :

Accepted Answer

We know V due to charge Q  =\dfrac {kQ}{r.} V is a scalar quantity \implies  total potential at origin is just sum of potential due to all charges 	herefore V=\dfrac {kq}{1}+\dfrac {kq}{2}+\dfrac {kq}{4}+....... =kq (\dfrac {1}{1-\dfrac {1}{2}})  (sum of infinite  G. P= \dfrac {a}{1-r}) 	herefore V=\dfrac {2Q}{4\pi \varepsilon_0}

Question 2

A rod with linear charge density  \lambda   is bent in the shape of circular ring. The electric potential at the center of the circular ring is :

Accepted Answer

let L be the length of nod \Rightarrow 2\pi R=L We know potential  =\dfrac{kQ}{R} Q=\lambda 	imes {total \ length} V=\dfrac{k\lambda 	imes L}{\left ( \dfrac{L}{2\pi } ight )}      =2\pi \lambda k And  k = \dfrac{1}{4 \pi \epsilon_{0}} V=\dfrac{\lambda}{2\epsilon_{0}}

Question 3

The potential at a point x (measured in  \mu m ) due to some charges situated on the x-axis is given by  V(x)=20/(x^{2}-4)  volt. The electric field  E  at  x = 4\mu m  is given by:

Accepted Answer

Given  V(x)=\dfrac{20}{x^{2}-4} We know,   E=-\dfrac{dV}{dx} \Rightarrow E=-\dfrac{d}{dx}(\dfrac{20}{x^{2}-4}) \Rightarrow E=-20\dfrac{-1(2x)}{(x^{2}-4)^{2}} \Rightarrow E=\dfrac{40x}{(x^{2}-4)^{2}} \Rightarrow E=\dfrac{40	imes4\ \mu m}{12	imes12} \Rightarrow E=\dfrac{10}{9} \ V/ \mu m

Question 4

Two identical small conducting spheres, having charges of opposite sign, attract each other with a force of  0.108N  when separated by  0.5m . The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of  0.036N . The initial charges of the spheres are :

Accepted Answer

Let initial charge be Q1 and Q2.Initial force   = F = \displaystyle\frac{K 	imes Q1 	imes Q2}{r^2} = -0.108       F = \displaystyle\frac{9 	imes 10^9 	imes Q1 	imes Q2}{0.25} = -0.108   Q1 \cdot Q2 = - 3 	imes 10^{-12}  Now on connecting them, since they are identical so they will have equal charges.Final charge   = \displaystyle\frac{Q1 + Q2}{2}  Final force   = F = \displaystyle\frac{9 	imes 10^9 	imes (Q1+Q2)^2 }{4 	imes 0.25} = 0.036   Q1 + Q2 = \pm 2 	imes 10^{-6}  Q2 = \pm 1.0	imes { 10 }^{ -6 }C\quad , Q1 = \mp 3.0	imes { 10 }^{ -6 }C

Question 5

An isolated metal sphere of radius  r  is given a charge  q . The potential energy of the sphere is :

Accepted Answer

Let at any instant  q  be the charge on the sphereNow potential of the sphere  =\dfrac {kq}{r} Work done to add a  dq  charge to this sphere dw=\dfrac {kq}{r}\cdot dq      (As  V_{\infty}=0 ) 	herefore   total work done to accumulate  q  charge is \int dw=\int_{0}^{q}\dfrac {kq}{r}\cdot dq \Rightarrow w=\dfrac {k}{r}[\dfrac {q^2}{2}]_{0}^{q} \Rightarrow w=\dfrac {kq^2}{2r} \Rightarrow w=\dfrac {q^2}{4\pi \varepsilon_0 r\cdot 2} \Rightarrow w=\dfrac {q^2}{8\pi \varepsilon_0r}

Question 6

Two point charges 2q and 8q are placed at a distance r apart. Where should a third charge -q be placed between them so that the electrical potential energy of the system is a minimum.

Accepted Answer

The potential energy can be as negative as we want by placing the negative charge infinitesimally close to one of the positive charges.

Question 7

A charge of  2\ C  is moved from a point  2\ m  away from a charge of  1\ C  to a point  1\ m  away from that charge. The work done is:

Accepted Answer

Voltage due to  1\ C  at a distance  r  is  \dfrac {k 	imes 1}{r} .Total change in potential in moving  2\ C  charge from  2\ m  away to  1\ m  away: V_2-V_1=\dfrac {k 	imes 1}{1}-\dfrac {k 	imes 1}{2} or,  V_2-V_1=\dfrac{k 	imes 1}{2} = \Delta V So, work done  =\Delta V Q =\dfrac {k 	imes 1}{2}	imes 2=9	imes 10^9\ J

Question 8

An electric dipole is along a uniform electric field. If it is deflected by  60^{o} , work done by the agent is  2 	imes 10^{-19}\ J . Then the work done by an agent if it is deflected by  30^{o}  further is:

Accepted Answer

If  P  be the dipole moment and  E  be the electric field then the potential energy of a dipole is given by: W= P.E. \cos  	heta Now, given:  P.E. \cos {60}^{o}  =  2 	imes {10}^{-19}\ J  Then,  P.E= 4 	imes {10}^{-19}\ J  Now, the work done to move the dipole further by   {30}^{o}  , W = P.E. \cos {90}^{o}  - P.E. \cos {60}^{o}  =0 - (4 	imes {10}^{-19} 	imes \frac{1}{2})\ J  or,  |W|=2 	imes {10}^{-19}\ J

Question 9

A solid non-conducting sphere of radius  R  having charge density  \rho = \rho_{0}x , where  x  is distance from the centre of sphere. The self potential energy of the sphere is

Accepted Answer

Given that,Non-conducting sphere, radius = RCharge density  =ho_0X ( x  is distance from the centre of sphere)To find self potential energy of sphere:- dE=\dfrac{Kdq}{x^2} dq=(ho_0x)(4\pi x^2)dx \displaystyle dE=\dfrac{1}{4\pi\in_0}	imes \dfrac{ho_0(4\pi x^2)}{x^2}xdx \displaystyle \int^E_0dE=\int^R_0\dfrac{ho_0}{\in_0}xdx  ( \because du=\dfrac{1}{2}(E)^2dv ) E=\dfrac{ho_0R^2}{2\in_0} \displaystyle du=\int^R_0\dfrac{ho^2_0R^4}{4\in^2_0}	imes \dfrac{1}{2}A\bar{x}x^2dx  ( \because dv=4\pi x^2dx ) =\dfrac{ho_0^2R^4\pi R^2}{3	imes 2} \boxed{U=\dfrac{\pi ho^2_0 R^6}{6\in_0}}

Question 10

Two circular plates of radius  R  are kept at distance  h  to make a parallel plate capacitor as shown in the figure. Then the capacitance of this capacitor is:

Accepted Answer

Capacitance of a parallel plate capacitor is:   C = {\varepsilon}_{o} \frac {A}{d}  , where  {\varepsilon}_{o}  is permittivity of free space,  A  is the surface area of each plate and  d  is the distance between the two plates.If  R  is the radius of the circular plate, then its area is given by  A = \pi R^{2} .Also given that  h  is the distance between the two plates.Hence,   C = \varepsilon_o  \dfrac {\pi R^{2}}{h} =  \dfrac {\varepsilon_{o} \pi R^{2}}{h}

Question 11

The ratio of the resultant capacities when three capacitors of  2 \mu F , 4 \mu F  and  6 \mu F  are connected first in series and then in parallel is

Accepted Answer

When capacitor in series \dfrac{1}{C_{eff_s}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}} \Rightarrow \dfrac{1}{C_{eff}}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6} \Rightarrow \dfrac{1}{C_{eff}}=\dfrac{6+3+2}{12} \Rightarrow{C_{eff}}=\dfrac{12}{11} C_{{eff}_{p}}=C_{1}+C_{2}+C_{3} =2+4+6 =12 	herefore \dfrac{C_{S}}{C_{P}}=\dfrac{1}{11}

Question 12

A number of identical condensers are first connected in parallel and then in series. The equivalent capacitance are found to be in the ratio  9:1 . The number of condensers used is :

Accepted Answer

In series \dfrac{1}{C_{eff}}=\dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C}+  upto n terms C_eff=\dfrac{C}{n}  In parallel C_eff=C+C+  upto n terms C_eff=nC \dfrac{C_{P}}{C_{S}}=\dfrac{9}{1} \dfrac{nC}{\dfrac{C}{n}}=\dfrac{9}{1} n^{2}=9 n=3

Question 13

The equivalent capacitance of the arrangement shown in figure, if A is the area of each plate is :

Accepted Answer

Hint: Capacitance of a capacitor is given by,  C = \dfrac{{{ \in _0}KA}}{d}  , where&#10;  K &#10;is the dielectric constant,  { \in _0} &#10;is permittivity of dielectric,  A &#10;is area of plate and  d  is&#10;the distance between the plates.&#10;&#10;Correct option: B&#10;&#10;Explanation for correct option:&#10;&#10;Step 1: Find capacitance of each capacitor.Capacitance of capacitor 1:  {C_1} = \dfrac{{{ \in _0}{K_1}\left( {\dfrac{A}{2}}&#10;\right)}}{d} = \dfrac{{{ \in _0}{K_1}A}}{{2d}} Capacitance of capacitor 2:  {C_2} = \dfrac{{{ \in _0}{K_2}\left( {\dfrac{A}{2}}&#10;\right)}}{{\left( {\dfrac{d}{2}} \right)}} = \dfrac{{{ \in _0}{K_2}A}}{d} Capacitance of capacitor 3:  {C_3} = \dfrac{{{ \in _0}{K_3}\left( {\dfrac{A}{2}}&#10;\right)}}{{\left( {\dfrac{d}{2}} \right)}} = \dfrac{{{ \in _0}{K_3}A}}{d} &#10;&#10;&#10;&#10;&#10;&#10;Step 2: Find the equivalent capacitance of the given&#10;circuit.&#10;&#10;Refer the given figure for the equivalent&#10;circuit.As  {C_2} &#10;and   {C_3} &#10;are in series and  {C_1}  is parallel&#10;to them, the equivalent capacitance will be:&#10;&#10;&#10;&#10; C = \dfrac{{{C_2}{C_3}}}{{{C_2}&#10;+ {C_3}}} + {C_1} &#10;&#10; &#10;\Rightarrow C = \dfrac{{\left( {\dfrac{{{ \in _0}{K_2}A}}{d}} \right)\left( {\dfrac{{{&#10;\in _0}{K_3}A}}{d}} \right)}}{{\left( {\dfrac{{{ \in _0}{K_2}A}}{d}} \right) +&#10;\left( {\dfrac{{{ \in _0}{K_3}A}}{d}} \right)}} + \dfrac{{{ \in&#10;_0}{K_1}A}}{{2d}} &#10;&#10; &#10;\Rightarrow C = \dfrac{{{ \in _0}A}}{d}\left( {\dfrac{{{K_1}}}{2} + \dfrac{{{K_2}{K_3}}}{{{K_2}&#10;+ {K_3}}}} \right) &#10;&#10;Hence, option B is correct.

Question 14

What is the capacitance of spherical capacitor having radius  a  and  b  where  (b>a) ?

Accepted Answer

It consists of two concentric spherical conducing shells of radii  a  and  b , say  b > a  (see fig.). The outer shell is earthed. Place a charge  +Q  on the inner shell. It will reside on the outer surface of the shell. A charge  -Q  will be induced on inner surface of outer shell. A charge  -Q  will be induced on inner surface of outer shell. A charge  +Q  will flow from outer shell to the earth.Consider a Gaussian spherical surface of radius  r  such that  a  a  is E = \dfrac{Q}{4\pi \varepsilon_0 r^2} Potential difference is V_b - V_a =\displaystyle -\int_a^b \vec{E} \cdot d\vec{r} = - \int_a^b \dfrac{q}{4\pi \varepsilon_0 r^2} dr Since  V_b = 0 , we have V_a = \int_a^b \dfrac{Q}{4\pi \varepsilon_0r^2} dr = \dfrac{Q}{4\pi \varepsilon_0}\left(\dfrac{1}{a} - \dfrac{1}{b}\right) = \dfrac{Q(b-a)}{4\pi \varepsilon _0 ab} Therefore, capacitance,  C = \dfrac{Q}{V_a - V_b} = \dfrac{Q}{V_a} = \dfrac{4\pi \varepsilon _0 ab}{b-a}

Electrostatic Potential And Capacitance