1638 + Interview Questions in ELECTROSTATICS IN GENERAL AWARENESS Page 20 InterviewSolution

951.	A semi-circular ring of mass m and radius R with linear charge ilensity `lambda`, hinged at its centre. is placed in a uniform electric field as shown in the figure `-1.451`. Ifthe ring is slightly rotated about O and released find the time period (in sec) of oscillation. Take `m = 8 kg, lambda= 2 C//m` and `E = 2 N//c`. Assume that coil is rotated in its own plane
Answer» Correct Answer - `6.28`

Discussion

952.	Calculate the voltage required to balanced an oil drop carrying 10 electrons, when located between plates of a capacitor, which are 5mm apart. Given mass of drop `= 3xx10^(-16) kg`, charge on electron `= 1.6xx10^(-19)C and g = 9.8 m//s^(2)`.
Answer» Correct Answer - 9.2V Here, `V = ?, n = 10, d = 5 mm = 5xx10^(-3) m`, `m = 3xx10^(-16) kg, e = 1.6xx10^(-19)C`, `g = 9.8m//s^(2)` To balance the oil drop `F = qE = mg or q((V)/(d)) = mg` `:. V = (mgd)/(q) = (mgd)/(n e) = (3xx10^(-16)xx9.8xx5xx10^(-3))/(10xx1.6xx10^(-19))` = 9.2 volt

Discussion

953.	A deuteron and an `alpha`-particle are placed in an electric field.The forces acting on them are `F_(1)` and `F_(2)` and their accelerations are `a_(1)` and `a_(2)` respectively. (i) `F_(1) = F_(2)` (ii) `F_(1) != F_(2)` (iii) `a_(1) = a_(2)` (iv) `a_(1) != a_(2)`A. `(i),(ii)`B. `(ii),(iii)`C. `(ii),(iv)`D. `(i),(iv)`
Answer» Correct Answer - 2 Deutron `(e,2m) , alpha`-particle `(2e , 4m)` `F_(1) = eE , F_(2) = 2eE` `a_(1) = (eE)/(2m) ,a_(2) = (2eE)/(4m) = (eE)/(2m)` `F_(1) != F_(2) ,a_(1) = a_(2)`

Discussion

954.	The linear charge density on upper half of semi-circular section of ring is `lambda` and that at lower half is `-lambda`. The direction of electric field at centre O of ring is : A. along OAB. along OBC. along OCD. along OD
Answer» Correct Answer - C

Discussion

955.	The magnitude of the electric field required to just balance in air a `2 xx 10^(-4) kg` liquid drop carrying a charge of `10 xx 10^(-2) mu C` isA. `10^(4) N//C`B. `2 xx 10^(4) N//C`C. `4 xx 10^(4) N//C`D. `5 xx 10^(4) N//C`
Answer» Correct Answer - 2 `qE = mg` `E = (mg)/(q) = ( 2 xx 10^(-4) xx 10)/(10 xx 10^(-2) xx 10^(-6)) = 2 xx 10^(4) N//C`

Discussion

956.	An electron of mass `m_(e )` initially at rest moves through a certain distance in a uniform electric field in time `t_(1)`. A proton of mass `m_(p)` also initially at rest takes time `t_(2)` to move through an equal distance in this uniform electric field.Neglecting the effect of gravity, the ratio of `t_(2)//t_(1)` is nearly equal toA. 1B. `sqrt(M_(p)/M_(e))`C. `sqrt(M_(e)/M_(p))`D. `1836`
Answer» Correct Answer - B

Discussion

957.	An electron of mass `m_(e )` initially at rest moves through a certain distance in a uniform electric field in time `t_(1)`. A proton of mass `m_(p)` also initially at rest takes time `t_(2)` to move through an equal distance in this uniform electric field.Neglecting the effect of gravity, the ratio of `t_(2)//t_(1)` is nearly equal toA. `1`B. `sqrt((m_(e ))/(m_(p)))`C. `sqrt((m_(p))/(m_(e)))`D. `1836`
Answer» Correct Answer - 3 `s = (1)/(2) a_(1) t_(1)^(2) = (1)/(2) a_(2) t_(2)^(2)` `(t_(2))/(t_(1)) = sqrt((a_(1))/(a_(2))) = sqrt((eE//m_(e))/(eE//m_(p))) = sqrt((m_(p))/(m_(e ))`

Discussion

958.	Electrons are caused to fall through a potential difference of 1500 V. If they are initially at rest, their final speed isA. `23xx10^(7)m//s`B. `2.3xx10^(7)m//s`C. `32xx10^(7)m//s`D. `3.2xx10^(7)m//s`
Answer» Correct Answer - B `v=sqrt((2qV)/m)=2.3xx10^(7)m//s`

Discussion

959.	If we use permittivity ` varepsilon` resistance `R`, gravitational constat `G` and voltage `V` as fundamental phusical quantities , then -.A. [angular displacement ]`= varepsilon^(1) R^(-2) G^(0) V^(0)`B. [velocity ]`== varepsilon^(1) R^(-2) G^(0) V^(0)`C. [dipole ]`= =varepsilon^(1) R^(-2) G^(0) V^(0)`D. [force ]`= =varepsilon^(1) R^(-2) G^(0) V^(0)`
Answer» Correct Answer - A::B::D

Discussion

960.	The radius of the earth is `6400km`, what is its capacitance?A. `1muF`B. `1mF`C. `1F`D. `10^(3)F`
Answer» Correct Answer - B

Discussion

961.	Explain the concept of electric field.
Answer» 1. The space around a charge gets modified when a test charge is brought in that region, it experiences a coulomb force. The region around a charged object in which coulomb force is experienced by another charge is called electric field. 2. Mathematically, electric field is defined as the force experienced per unit charge. 3. The coulomb force acts across an empty space (vacuum) and does not need any intervening medium for its transmission. 4. The electric field exists around a charge irrespective of the presence of other charges. 5. Since the coulomb force is a vector, the electric field of a charge is also a vector and is directed along the direction of the coulomb force, experienced by a test charge.

961.

Explain the concept of electric field.

Answer»

1. The space around a charge gets modified when a test charge is brought in that region, it experiences a coulomb force.

The region around a charged object in which coulomb force is experienced by another charge is called electric field.

2. Mathematically, electric field is defined as the force experienced per unit charge.

3. The coulomb force acts across an empty space (vacuum) and does not need any intervening medium for its transmission.

4. The electric field exists around a charge irrespective of the presence of other charges.

5. Since the coulomb force is a vector, the electric field of a charge is also a vector and is directed along the direction of the coulomb force, experienced by a test charge.

Discussion

962.	Find the potential difference `V_(AB)` between `A (0,0,0) and B (1 m , 1 m,1 m)` in an electric field : (i) `vec E = (y hat i + x hat j) Vm^-1` (ii) `vec E = (3 x^2 y hat i + x^3 hat j) Vm^-1`.
Answer» (a) `dV = -(E_(x) dx + E_(y) dy) = -(ydx + xdy)` `int y dx = yx` or `int xdy = xy` Since integer of both terms is same i.e. `vec(E ) . D vec(r )` is a perfect differential. `V_(1,2,3) - V_(0,0,0) = - int_(0,0,0)^(1,2,3) vec(E ) . d vec(r ) = - int(E_(x) dx + E_(y) dy)` `= - int (y dx + xdy) = - int d(xy) = -xy` `= -\|xy\|_(0 ,0)^(1,2)` `= - [ (1 xx 2) - (0 , 0)] = -2` `= -2` volt (b) `dV = -(E_(x) dx + E_(y) dx)` ` = -(3 x^(2) y dx + x^(3) dy)` `int3x^(2) y dx = y int 3 x^(2) dx = y . 3 (x^(3))/(3) = x^(3) y` `int x^(3) dy = x^(3) int dy = x^(3) y` Since integral of both terms is same , i.e. `vec(E ) . d vec(r)` is perfect differential. `V_(1,2,3) - V_(0,0,0) = - int_(0,0,0)^(1,2,3) (3x^(2) y dx + x^(3) dy) = - int_(0,0,0)^(1,2,3) d(x^(3) y)` `= - [x^(3) y ]_(0,0)^(1,2) = - [(1^(3) xx 2) - 0 xx 0)]` ` = - 2` volt

Discussion

963.	Charge on an electron is 1.6 × 10-19 C. How many electrons are required to accumulate a charge of one coulomb?
Answer» 1 electron = 1.6 × 10^-19 C ∴ 1 C = \(\frac{1}{1.6\times10^{-19}}\) electrons = 0.625 × 10¹⁹electrons ……. (Taking reciprocal from log table) = 6.25 × 10¹⁸electrons Hence, 6.25 × 10¹⁸ electrons are required to accumulate a charge of one coulomb.

963.

Charge on an electron is 1.6 × 10-19 C. How many electrons are required to accumulate a charge of one coulomb?

Answer»

1 electron = 1.6 × 10^-19 C

∴ 1 C = \(\frac{1}{1.6\times10^{-19}}\) electrons

= 0.625 × 10¹⁹electrons

……. (Taking reciprocal from log table)

= 6.25 × 10¹⁸electrons

Hence, 6.25 × 10¹⁸ electrons are required to accumulate a charge of one coulomb.

Discussion

964.	`A` and `B` are two points in an electric field . If the work done in carrying `4.0 C` of electric charge from `A` to `B` is `16.0 J`, the potential difference between `A` and `B` isA. zeroB. `2.0 V`C. `4.0 V`D. `16.0 V`
Answer» Correct Answer - 3 `Delta V = (W)/(q) = (16)/(4) = 4 V`

Discussion

965.	STATEMENT -1: A conducting sphere charged upto ` 50V` is placed at the centre of a conducting shell charged upto ` 100V` and connected by a wire . All the charge of the shell flows to the sphere. STATEMENT -2: The positive charge always flows from higher to lower potential .A. Statement -1 is True , Statement -2 is True , Statement -2 is a correct explanation for Statement -1B. Statement -1 is True , Statement -2 is True , Statement -2 is NOT a correct explanation for statement -1.C. Statement -1 true, Statement -2 is False.D. Statement -1 is False , Statement -2 is true.
Answer» Correct Answer - D

Discussion

966.	STATEMENT -1: A point charge `q` is placed in front of a solid conducting sphere. Electric field due to induced charges at the centre of sphere is zero . . STATEMNT -2 : Electric field at a point inside the solid body fo conductor is zero.A. Statement -1 is True , Statement -2 is True , Statement -2 is a correct explanation for Statement -9B. Statement -1 is Tre , Statement -2 is True , Statement -2 is NOT a correct explanation for statement -1.C. Statement -1 true, Statement -2 is False.D. Statement -1 is False , Statement -2 is true.
Answer» Correct Answer - D

Discussion

967.	In the circuit as shown in the figure the effective capacitance between `A` and `B` is
Answer» Correct Answer - 4

Discussion

968.	In the given figure, a diode `D` is connected to an external resistance `R=100 Omega` and an emf of `3.5 V`. If the barrier potential developed across the diode is `0.5 V`, the current in the circuit will be : A. `30mA`B. `40mA`C. `20mA`D. `35mA`
Answer» Correct Answer - A

Discussion

969.	A plane surface of element of area `1 mm^(2)` is situated in a uniform electric field of intensity `9xx10^(6)N//C` with its plane making an angle of `30^(@)` with the direction of the field. The electric flux through the surface element isA. `4.5Nm^(2)//C`B. `3.5Nm^(2)//C`C. `0.5Nm^(2)//C`D. `2.5Nm^(2)//C`
Answer» Correct Answer - A `phi=dSE" ""as"" "theta=1xx10^(-6)xx9xx10^(6)xxcos60` `=4.5Nm^(2)//C.`

Discussion

970.	The electric potential at the surface of an atomic nucleus (Z = 50) of radius `9xx10^(-15)m` isA. `4xx10^(6)V`B. `8xx10^(6)V`C. `4xx10^(-6)V`D. `8xx10^(-6)V`
Answer» Correct Answer - D `V=1/(4piin_(0))q/r=1/(4piin_(0))(Ze)/r=8xx10^(6)V`

Discussion

971.	The four capacitors, each of `25 muF` are connected as shown in figure. The DC voltmeter reads 200V. The charge on each plate of capacitor is A. `2xx10^(-3)C`B. `5xx10^(-3)C`C. `2xx10^(-2)C`D. `5xx10^(-2)C`
Answer» Correct Answer - B All the capacitors are in parallel and potential difference across each is 200V. `therefore" "Q=CV=25xx10^(-6)xx200=5xx10^(-3)C`

Discussion

972.	A metal sphere of radius 10 cm is charged to a potential of 100 V. The outward pull per unit area of the surface isA. `4.4xx10^(-6)N//m^(2)`B. `2.4xx10^(-6)N//m^(2)`C. `4.4xx10^(6)N//m^(2)`D. `4.2xx10^(-6)N//m^(2)`
Answer» Correct Answer - A `f=1/2epsi_(0)k(V/d)^(2)` `=(8.85xx10^(-12)xx1)/2xx(100/(10xx10^(-2)))^(2)` `=4.4xx10^(-6)N//m^(2).`

Discussion

973.	The figure shown represents the electric field between two large metal plates. Which of the graphs given in figure represents the force on a positve charge as it moves from point P towards plate B? A. B. C. D.
Answer» Correct Answer - D Electric field produced by large metal plate does not vary with distance so force on charge will be constant.

Discussion

974.	Calculate the equuivalent capacitances between the points `A` and `B` in the combination shown in Fig.
Answer» Correct Answer - `13.44 mu F` In Fig, `8 mu F` and `10 mu F` capacitors are in series and the combination is in parallel with `9 mu F` capacitor.

Discussion

975.	Two capacitors of `2 muF` and `3 mu F` are joined in series. The outer plate of second capacitor is earthed. Find out the potential and charge of the inner plate of each capacitor.
Answer» Correct Answer - `400 V, 400 V , 1.2xx10^(-3) C` We find `V_(1) = 600` volt and `v_(2) = 400` volt. As outer plate of first capacitor is at `1000`volt, its inner plate must be at `1000-600 = 400` volt Further, as outer plate of second capacitor is earthed , i.e., at zero potential, therefore inner plate of second capacitor must be at `400-0 = 400` volt Charge on inner plate of each condenser `q = C_(1) V_(1) = (2xx10^(-6))xx600 = 1.2xx10^(-3) C`

Discussion

976.	A positive charge q is placed in front of conducting solid cube at a distance d from its centre. Find the electric field at the centre of the cube due to the charges appearing on its surface.A. `(1)/(4 pi in_(0)) .(q)/(d^(2))`B. `(1)/(4 pi in_(0)) .(2q)/(d^(2))`C. `(1)/(4 pi in_(0)) .(3q)/(d^(2))`D. `(1)/(4 pi in_(0)) .(4q)/(d^(2))`
Answer» Correct Answer - 1

Discussion

977.	In a certain charge distribution, all points having zero potetnial can be joined by a circles `S`. Points inside `S` have positive potential and points outside `S` have negative potential. A positive charg, which is free to move, is placed inside `S`A. It will remainin equilibriumB. It can move inside `S`, but it cannot cross `S`C. It must cross `S` at some time.D. It may move, but will ultimately return to its starting point.
Answer» Correct Answer - C

Discussion

978.	A spherical equipotential surface is not possibleA. for a point charggeB. for a dipoleC. inside a uniformly charged sphereD. inside a spherical capacitor
Answer» Correct Answer - B

Discussion

979.	The electric field due to uniformly charged sphere of radius `R` as a function of the distance from its centre is represented graphically byA. B. C. D.
Answer» Correct Answer - 2

Discussion

980.	As isolated sphere has a capacitance `60 pF`. (i) Calculate its radius. (ii) How much charge should be placed on it to raise its potential to `10^(4) V` ?
Answer» Here, `C = 60 pF = 60 xx 10^(-12) F`, `V = 10^(4)V` (i) `r = (1)/(4pi in_(0)) C = 9xx10^(9)xx60xx10^(-12)` `= 54xx10^(-2)m = 54m` (ii) `q = CV = 60xx10^(-12)xx10^(4)` `= 60xx10^(-8) = 0.6 muC`

Discussion

981.	If the capacitance of a conductor carrying a charge of 58C is 0.05 F, calculate its potential.
Answer» Here, C = 0.005 F, Q = 8 C,V= ? `V = (Q)/(C ) = (8)/(0.005) = 1600 V`

Discussion

982.	A charge of `+2.0xx 10^(-8)` C is placed on the positive place and a charge of `-1.0 xx 10^(-8) C` on the negative plate of a parallel- plate capacitor of capacitance `1.2 xx 10^(-3) mu` F. Calculate the potential difference developed between the plates.
Answer» Here, `q_(1) = 2.0xx10^(-8) C`. `q_(2) = 1.0xx10^(-8) C` `C = 1.2xx10^(-3) muF = 1.2xx10^(-9) F, V = ?` `V = (q_(1) - q_(2))/(2C) = (2.0xx10^(-8) -(-1.0xx10^(-8)))/(2xx1.2xx10^(-9))` 12.5V

Discussion

983.	When `1.0 xx 10^12`electrons are transferred from one conductor to another, a potential difference of `10 V` find the capacitance of the two -conductor system .
Answer» Here, charge transferred, Q =n e `Q = (1.0xx10^(12)) (1.6xx10^(-19)) C` `= 1.6xx10^(-7)C` , V = 10 Volt, C = ? As `C = (Q)/(V) = (1.6xx10^(-7))/(10) = 1.6xx10^(-8) F`

Discussion

984.	At a certain distance from a point charge, the field intensity is `500 V//m` and the potentil is `-3000 V`. The distance to the charge and the magnitude of the charge respectively areA. `6m` and `6muC`B. `4m` and `2muC`C. `6m` and `4muC`D. `6m` and `2muC`
Answer» Correct Answer - D `500=(k\|q\|)/r^2`…………i `-3000=(k(-q))/r`……..ii Solving these two equations we get `r=6m` `:. \|q\|=(500r^2)/(k) `=((500)(6)^(2))/((9xx10^9)` `=2xx10^-6C` `=2muC`

Discussion

985.	Two points charges `q_1 and q_2` are placed at a distance of 50 m from each each other in air, and interact with a certain force. The same charges are now put in oil whose relative permittivity is 5. If the interacting force between them is still the same, their separation now isA. `16.6m`B. `22.3m`C. `28.4m`D. `25.0m`
Answer» Correct Answer - B `1/(4piepsilon_0).(q_1q_2)/r_1^2=1/(4piepsilon_0k).(q_1q_2)/r_2^2)` `:. r_2=r_1/sqrtk=50/sqrt5` `=22.36m`

Discussion

986.	An electrical charge `2xx10^-8` C is placed at the point `(1,2,4)` m. At the point `(4,2,0) `m, the electricA. potential will be 36 VB. field will along y-axisC. field will increase if the space between the points is filled with a dielectricD. All of the above
Answer» Correct Answer - A `r=sqrt((4-1)^2+(2-2)^2+(0-4)^2)` `=5m` `V=1/(4piepsilon).q/r` `=((9xx10^9)(2xx10^-8))/5=36V` Field is in the direction of `r=r_p-r_q`

Discussion

987.	A uniform electric field of strength E exists in region. A electron enters a point A with velocity v as shown. It moves through the electric field and reaches at point B. Velocity particle at B is `2v` at `30^@` with x-axis . Then A. electric field `E=(3mv^2)/(2ea) hati`B. rate of doing work doen by electric field ast B is `(3mv^2)/(2ea)`C. Bota a and b are correctD. both a and b are wrong
Answer» Correct Answer - A `(v_A)y=vimplies(v_B)_y=2vsin30^@v` Since y-component of velocity remains unchanged. Hence electric field is along `(-hati)` direction. Work done by electrostatic force in moving from A and B =change in its kinetic energy `:. (eE)(2a-a)1/2m(4v^2-v^2)` `E=(mv^2)/(2ea)` `or E=-(3mv^2)/(2ea)hati` Rate of doing work done =power `=Fvcostheta` `((3mv^2)/(2ea))^2(e)(2v)cos30^@` `=3sqrt3/2(mv^2)/a`

Discussion

988.	Where is the knowledge of dielectric strength helpful ?
Answer» Dielectric strength is the maximum strength of electric field that can be tolerated by the dielectric without electric breakdown its knowledge helps us in designing a capacitor by determining the maximum potential difference that can be applied across the plates of the capacitor.

Discussion

989.	The inroduction fo dielectric slab between the capacitor plates increases the capacitance. Why ?
Answer» A dielectric slab of dielectric constant K reduces the electric field from E to `E//K`. This reduces the potential differences from V to `V//K`. Hence the capacitance increases from C to KC.

Discussion

990.	Protons are projected with an initial speed `v_i = 9.55 xx 10^3 m//s` into a region where a uniform electric field `E = (-720hat j`) N/C is present, as shown in figure. The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons are launched. Find (a) the two projection angles 0 that result in a hit and (b) the total time of flight for each trajectory.
Answer» Correct Answer - A::B a. `g_e=g+(qE)/m` `=10+((1.6xx10^-19)(720))/(1.67xx10^-27)` `~~6.9xx10^10m//s^2` Now, `R=(u^2sin2theta)/g_e` `:. (1.27xx10^-3)=((9.55xx10^3)^2sin2theta)/(6.9xx10^10)` Solving this equation we get `theta=37^@ and 53^@` b. Applying `T=(2usintheta)/g_e`

Discussion

991.	The electric field intensity at a point 2 m from an isolated point charge is 500 N/C. The electric potential at the point is(A) 0 V (B) 2.5 V (C) 250 V (D) 1000 V
Answer» Correct option is: (D) 1000 V

Discussion

992.	What is the magnitude of charge on an electron?
Answer» The magnitude of charge on an electron is 1.6 × 10^-19 C.

Discussion

993.	Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be(A) 1.0 N (B) 9 × 109 N (C) 9 × 10-9 N (D) 10 N
Answer» Correct answer is (B) 9 × 10⁹ N

Discussion

994.	State the law of conservation of charge.
Answer» In any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant” OR For an isolated system, total charge cannot be created nor destroyed.

994.

State the law of conservation of charge.

Answer»

In any given physical process, charge may get transferred from one part of the system to another, but the total charge in the system remains constant”

OR

For an isolated system, total charge cannot be created nor destroyed.

Discussion

995.	Two point charges of +5 µC are so placed that they experience a force of 80 × 10-3 N. They are then moved apart, so that the force is now 2.0 × 10-3 N. The distance between them is now (A) 1/4 the previous distance (B) double the previous distance (C) four times the previous distance (D) half the previous distance
Answer» Correct answer is (B) double the previous distance

Discussion

996.	Define a unit charge.
Answer» Unit charge (one coulomb) is the amount of charge which, when placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of 9.0 × 10⁹ N.

Discussion

997.	State the units of linear charge density.
Answer» SI unit of λ is (C / m).

Discussion

998.	If two lines of force intersect of one point. What does it mean?
Answer» If two lines of force intersect of one point, it would mean that electric field has two directions at a single point.

Discussion

999.	Three equal charges are placed on the three corners of a square. If the force between `q_(1)` and `q_(2)` is `F_(12)` and that between `q_(1)` and `q_(3)` is `F_(13)`, then the ratio of magnitudes `(F_(12)//F_(13))` isA. `1//2`B. 2C. `1//sqrt2`D. `sqrt2`
Answer» Correct Answer - B

Discussion

1000.	The insulation property of air breaks down at `E=3xx10^(6) "volt"//meter`. The maximum charge that can be given to a sphere of diameter `5m` is approximately (in coulombs)A. `2 xx 10^(-2)`B. `2 xx 10^(-3)`C. `2 xx 10^(-4)`D. `2 xx 10^(-5)`
Answer» Correct Answer - B

Discussion

Explore topic-wise InterviewSolutions in .

Calculate the voltage required to balanced an oil drop carrying 10 electrons, when located between plates of a capacitor, which are 5mm apart. Given mass of drop `= 3xx10^(-16) kg`, charge on electron `= 1.6xx10^(-19)C and g = 9.8 m//s^(2)`.

The linear charge density on upper half of semi-circular section of ring is `lambda` and that at lower half is `-lambda`. The direction of electric field at centre O of ring is : A. along OAB. along OBC. along OCD. along OD

The magnitude of the electric field required to just balance in air a `2 xx 10^(-4) kg` liquid drop carrying a charge of `10 xx 10^(-2) mu C` isA. `10^(4) N//C`B. `2 xx 10^(4) N//C`C. `4 xx 10^(4) N//C`D. `5 xx 10^(4) N//C`

Electrons are caused to fall through a potential difference of 1500 V. If they are initially at rest, their final speed isA. `23xx10^(7)m//s`B. `2.3xx10^(7)m//s`C. `32xx10^(7)m//s`D. `3.2xx10^(7)m//s`

The radius of the earth is `6400km`, what is its capacitance?A. `1muF`B. `1mF`C. `1F`D. `10^(3)F`

Explain the concept of electric field.

Find the potential difference `V_(AB)` between `A (0,0,0) and B (1 m , 1 m,1 m)` in an electric field : (i) `vec E = (y hat i + x hat j) Vm^-1` (ii) `vec E = (3 x^2 y hat i + x^3 hat j) Vm^-1`.

Charge on an electron is 1.6 × 10-19 C. How many electrons are required to accumulate a charge of one coulomb?

`A` and `B` are two points in an electric field . If the work done in carrying `4.0 C` of electric charge from `A` to `B` is `16.0 J`, the potential difference between `A` and `B` isA. zeroB. `2.0 V`C. `4.0 V`D. `16.0 V`

In the circuit as shown in the figure the effective capacitance between `A` and `B` is

In the given figure, a diode `D` is connected to an external resistance `R=100 Omega` and an emf of `3.5 V`. If the barrier potential developed across the diode is `0.5 V`, the current in the circuit will be : A. `30mA`B. `40mA`C. `20mA`D. `35mA`

The electric potential at the surface of an atomic nucleus (Z = 50) of radius `9xx10^(-15)m` isA. `4xx10^(6)V`B. `8xx10^(6)V`C. `4xx10^(-6)V`D. `8xx10^(-6)V`

The four capacitors, each of `25 muF` are connected as shown in figure. The DC voltmeter reads 200V. The charge on each plate of capacitor is A. `2xx10^(-3)C`B. `5xx10^(-3)C`C. `2xx10^(-2)C`D. `5xx10^(-2)C`

A metal sphere of radius 10 cm is charged to a potential of 100 V. The outward pull per unit area of the surface isA. `4.4xx10^(-6)N//m^(2)`B. `2.4xx10^(-6)N//m^(2)`C. `4.4xx10^(6)N//m^(2)`D. `4.2xx10^(-6)N//m^(2)`

The figure shown represents the electric field between two large metal plates. Which of the graphs given in figure represents the force on a positve charge as it moves from point P towards plate B? A. B. C. D.

Calculate the equuivalent capacitances between the points `A` and `B` in the combination shown in Fig.

Two capacitors of `2 muF` and `3 mu F` are joined in series. The outer plate of second capacitor is earthed. Find out the potential and charge of the inner plate of each capacitor.

A spherical equipotential surface is not possibleA. for a point charggeB. for a dipoleC. inside a uniformly charged sphereD. inside a spherical capacitor

The electric field due to uniformly charged sphere of radius `R` as a function of the distance from its centre is represented graphically byA. B. C. D.

As isolated sphere has a capacitance `60 pF`. (i) Calculate its radius. (ii) How much charge should be placed on it to raise its potential to `10^(4) V` ?

If the capacitance of a conductor carrying a charge of 58C is 0.05 F, calculate its potential.

A charge of `+2.0xx 10^(-8)` C is placed on the positive place and a charge of `-1.0 xx 10^(-8) C` on the negative plate of a parallel- plate capacitor of capacitance `1.2 xx 10^(-3) mu` F. Calculate the potential difference developed between the plates.

When `1.0 xx 10^12`electrons are transferred from one conductor to another, a potential difference of `10 V` find the capacitance of the two -conductor system .

At a certain distance from a point charge, the field intensity is `500 V//m` and the potentil is `-3000 V`. The distance to the charge and the magnitude of the charge respectively areA. `6m` and `6muC`B. `4m` and `2muC`C. `6m` and `4muC`D. `6m` and `2muC`

An electrical charge `2xx10^-8` C is placed at the point `(1,2,4)` m. At the point `(4,2,0) `m, the electricA. potential will be 36 VB. field will along y-axisC. field will increase if the space between the points is filled with a dielectricD. All of the above

Where is the knowledge of dielectric strength helpful ?

The inroduction fo dielectric slab between the capacitor plates increases the capacitance. Why ?

The electric field intensity at a point 2 m from an isolated point charge is 500 N/C. The electric potential at the point is(A) 0 V (B) 2.5 V (C) 250 V (D) 1000 V

What is the magnitude of charge on an electron?

Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be(A) 1.0 N (B) 9 × 109 N (C) 9 × 10-9 N (D) 10 N

State the law of conservation of charge.

Define a unit charge.

State the units of linear charge density.

If two lines of force intersect of one point. What does it mean?

Three equal charges are placed on the three corners of a square. If the force between `q_(1)` and `q_(2)` is `F_(12)` and that between `q_(1)` and `q_(3)` is `F_(13)`, then the ratio of magnitudes `(F_(12)//F_(13))` isA. `1//2`B. 2C. `1//sqrt2`D. `sqrt2`

The insulation property of air breaks down at `E=3xx10^(6) "volt"//meter`. The maximum charge that can be given to a sphere of diameter `5m` is approximately (in coulombs)A. `2 xx 10^(-2)`B. `2 xx 10^(-3)`C. `2 xx 10^(-4)`D. `2 xx 10^(-5)`