Q: A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

\textbf{Step 1 : Formula of Energy stored in capacitor} The Energy stored U in a capacitor of Capacitance C , Charge Q and Potential Difference V across it is given by: U = \cfrac 12 CV^2 = \cfrac {Q^2}{2C} = \cfrac 12 QV ....(1) \textbf{Step 2: Calculation of Energy stored} Given, Capacitance C = 12pF = 12 \times 10^{-12} F And Potential difference V= 50V So using the first formula from Equation (1) , we get Energy stored as: U=\dfrac{1}{2}CV^2=\dfrac{1}{2}\times [12\times 10^{-12}F]\times [(50 V)^2]=15\times 10^{-9}\,J \boxed{U=15\,nJ} Hence Energy stored in capacitor is 15\,nJ .

Q: An electric dipole of length 4cm when placed with its axis making an angle of {60^ \circ } with a uniform electric field, experience a torque of 4\sqrt 3 N - m . Calculate the potential energy of the dipole of it has charge \pm 8nC .

Dipole moment(P) = q.2L P=8\times { 10 }^{ -9 }\times 2\times 4\times { 10 }^{ -2 }\\ \quad =64\times { 10 }^{ -11 }Cm Torque on the dipole in uniform electric field \tau =PEsin\theta \\ E=\dfrac { \tau }{ Psin\theta } Potential energy of the dipole is U=-PEcos\theta on substituting the value , we get the value of E and U U=-P\dfrac { \tau }{ Psin\theta } cos\theta \\ \quad \quad =\quad -\dfrac { \tau }{ tan\theta } \quad U\quad =-\dfrac { 4\sqrt { 3 } }{ \sqrt { 3 } } =-4Nm

Q: Determine the potential difference across the plates of the capacitor 'C_1' of the network shown in the figure. [Assume E_2 > E_1 ]

b'As E_2>E_1 \therefore the direction is clockwise using kirchoff rule, \dfrac{q}{C_2}-E_1+ \dfrac{q}{C_1}+E_2=0 as V_1 =\dfrac{q}{C_1} \ and \ V_2 =\dfrac{q}{C_2 } \therefore V_1 = E_1 - E_2 -\dfrac{q}{C_2} V_2 = E_1 -E_2- \dfrac{q}{C_1} '

Q: In the circuit shown, a time varying voltage V=2000t volt is applied where t is in second. At time t=5 ms . determine the current through the resistor R=4\Omega and through the capacitor C=300\mu F .

At t=5 ms, V=10 V \therefore i_R=\dfrac {V}{R}=\dfrac {10}{4}=2.5 A Further q=CV=(300\times 10^{-6})(2000t)=0.6t i_C=\dfrac {dq}{dt}=0.6A=constant

Q: Two point charges 5 \times 10^{-8} C and -3 \times 10^{-8} C are located 16 cm apart. At what point on the line joining these charges the electric potential will be zero?

Let the distance of the point from charge 5\times 10^{-8}C be r .Hence the distance of this point from the other charge is 16-r .Since the potential at the point is zero, \dfrac{k(5\times 10^{-8})}{r}+\dfrac{k(-3\times 10^{-8})}{16-r}=0 \implies r=10cm

Question 1

Define capacitance of a capacitor and its SI unit.

Accepted Answer

The capacitance of any capacitor is its ability to store charge  C=\dfrac{Q}{V} C=  capacitance  Q=  Charge  V=  Potential across capacitor.The SI unit of capacitance is farad(F).

Question 2

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Accepted Answer

extbf{Step 1 : Formula of Energy stored in capacitor} The Energy stored  U  in a capacitor of Capacitance  C , Charge  Q  and Potential Difference  V  across it is given by:             U = \cfrac 12 CV^2 = \cfrac {Q^2}{2C} = \cfrac 12 QV                 ....(1) 	extbf{Step 2: Calculation of Energy stored} Given, Capacitance  C = 12pF = 12 	imes 10^{-12} F And Potential difference  V= 50V So using the first formula from Equation  (1) , we get Energy stored as:                  U=\dfrac{1}{2}CV^2=\dfrac{1}{2}	imes [12	imes 10^{-12}F]	imes [(50 V)^2]=15	imes 10^{-9}\,J                   \boxed{U=15\,nJ} Hence Energy stored in capacitor is  15\,nJ .

Question 3

Define electric potential.

Accepted Answer

The electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point.You may wonder what the hell is infinity. It is not practically possible.The concept of absolute electric potential is only theoretical.What practically exists is the potential difference between two points. It is defined as the work done in moving a unit charge from one point to another.In the theory the term infinity arises because only at infinity we can consider that the electric potential is absolutely zero.

Question 4

A parallel-plate air capacitor of capacitance  245 pF  has a charge of magnitude  0.148\mu C  on each plate. The plates are  0.328 mm  apart. What is the area of each plate?

Accepted Answer

The capacitance of parallel plate air capacitor is  C=\dfrac{A\epsilon_0}{d} So, area of each plate  A=\dfrac{Cd}{\epsilon_0}=\dfrac{(245	imes 10^{-12})(0.328	imes 10^{-3})}{8.85	imes 10^{-12}}=9.08 	imes 10^{-3}m^2=90.8 cm^2

Question 5

An electric dipole of length 4cm when placed with its axis making an angle of  {60^ \circ }  with a uniform electric field, experience a torque of  4\sqrt 3 N - m . Calculate the potential energy of the dipole of it has charge   \pm 8nC .

Accepted Answer

Dipole moment(P) = q.2L P=8	imes { 10 }^{ -9 }	imes 2	imes 4	imes { 10 }^{ -2 }\ \quad =64	imes { 10 }^{ -11 }Cm Torque on the dipole in uniform electric field 	au =PEsin	heta \ E=\dfrac { 	au  }{ Psin	heta  }  Potential energy of the dipole is  U=-PEcos	heta  on substituting the value , we get the value of E and U U=-P\dfrac { 	au  }{ Psin	heta  } cos	heta \ \quad \quad =\quad -\dfrac { 	au  }{ tan	heta  }  \quad U\quad =-\dfrac { 4\sqrt { 3 }  }{ \sqrt { 3 }  } =-4Nm

Question 6

Determine the potential difference across the plates of the capacitor  'C_1'  of the network shown in the figure. [Assume  E_2 > E_1 ]

Accepted Answer

b'As  E_2>E_1   	herefore  the direction is clockwise using kirchoff rule,  \dfrac{q}{C_2}-E_1+ \dfrac{q}{C_1}+E_2=0 as  V_1 =\dfrac{q}{C_1} \ and \ V_2 =\dfrac{q}{C_2 } 	herefore V_1 = E_1 - E_2 -\dfrac{q}{C_2}     V_2 = E_1 -E_2- \dfrac{q}{C_1}   '

Question 7

A  4.00\mu F  capacitor and a  6.00\mu F  capacitor are connected in parallel across a  660 V  supply line. Find the charge on each capacitor and the voltage across each of them.

Accepted Answer

As the both capacitor are in parallel so potential across them will be same i.e  660 V .Now using  Q=CV , the charge on  4\mu F  is  Q_1=(4	imes 10^{-6})	imes 660=2.64	imes 10^{-3} C  and the charge on  6\mu F  is  Q_2=(6	imes 10^{-6})	imes 660=3.96	imes 10^{-3} C

Question 8

In the circuit shown, a time varying voltage

V = 2000 t

volt is applied where t is in second. At time

t = 5 m s

. determine the current through the resistor

R = 4 Ω

and through the capacitor

C = 300 μ F

.

Accepted Answer

At  t=5 ms, V=10 V 	herefore i_R=\dfrac {V}{R}=\dfrac {10}{4}=2.5 A Further  q=CV=(300	imes 10^{-6})(2000t)=0.6t i_C=\dfrac {dq}{dt}=0.6A=constant

Question 9

Two point charges  5 	imes 10^{-8} C  and  -3 	imes 10^{-8} C  are located 16 cm apart. At what point on the line joining these charges the electric potential will be zero?

Accepted Answer

Let the distance of the point from charge  5	imes 10^{-8}C  be  r .Hence the distance of this point from the other charge is  16-r .Since the potential at the point is zero, \dfrac{k(5	imes 10^{-8})}{r}+\dfrac{k(-3	imes 10^{-8})}{16-r}=0 \implies r=10cm

Question 10

Question 11

The initial and final stored energy in the capacitor,  U_i  and  U_f ?

Accepted Answer

The initial parallel plate capacitance is  C_i=\dfrac{A\epsilon_0}{d} and the charge is  Q_i=C_iV_i=\dfrac{A\epsilon_0 V}{d} The final parallel plate capacitance is  C_f=\dfrac{A\epsilon_0}{2d} and charge is   Q_f=C_fV_f=\dfrac{A\epsilon_0 V_f}{2d} As the battery is disconnected so the charge will remain constant so, Q_i=Q_f or  \dfrac{A\epsilon_0 V}{d}=\dfrac{A\epsilon_0 V_f}{2d} or,  V_f=2V Thus, initial energy stored,  U_i=\dfrac{1}{2}C_iV_i^2=\dfrac{A\epsilon_0 V^2}{2d}  and final energy stored,  U_f=\dfrac{1}{2}C_fV_f^2=\dfrac{A\epsilon_0 (2V)^2}{4d}=\dfrac{A\epsilon_0}{d}V^2

Question 12

A test charge  'q'  is moved without acceleration from  A  to  C  along the path from  A  to  B  and then from  B  to  C  in electric field  E  as shown in the figure. (i) Calculate the potential difference between  A  and  C . (ii) At which point (of the two) is the electric potential more and why?

Accepted Answer

Hint: Use the formula of potential difference between two points in uniform electric field.Explanation:- 	extbf{Step1: Calculate potential difference between points} (i)Calculate potential difference between two points  V_C - V_A = -E x    V_C - V_A = -E (2-6) =4E 	extbf{Step2: Find at which point electric potential is high} (ii)Electric potential is more at point C as electric field is always directed from higher potential region to lower potential region.

Electrostatic Potential And Capacitance