InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Across the surface of a charged conductor, the electricA. field is continuousB. potential is continuousC. field is discontinousD. potential is discontinuous |
|
Answer» Correct Answer - B::C |
|
| 102. |
Four plates of the same area of cross-section are joined as shown in the figure. The distance each plate is `d`. The equivalent capacity across `A` and `B` will be A. `(2epsilon_(0)A)/(d)`B. `(3epsilon_(0)A)/(d)`C. `(3epsilon_(0)A)/(2d)`D. `(epsilon_(0)A)/(d)` |
|
Answer» Correct Answer - B The given arrangement is equivalent to the parallel combination of three identical capacitors. Hence equivalent capacitance `= 3C = 3 (epsilon_(0)A)/(d)` |
|
| 103. |
Three plates `A, B, C` each of area `50 cm^(2)` have separation `3 mm` between `A` and `B mm` between `B` and `C`. The energy stored when the plates area fully charged is A. `1.6 xx 10^(-9) J`B. `2.1 xx 10^(-9) J`C. `5 xx 10^(-9) J`D. `7 xx 10^(-9) J` |
|
Answer» Correct Answer - B There are two capacitors parallel to each other. `:.` Total capacitance `= (2epsilon_(0)A)/(d)` `:.` Energy stored ` = (1)/(2) ((2epsilon_(0)A)/(d)) V^(2)` `= (8.86 xx 10^(-12) xx 50 xx 10^(-4) xx 12^(2))/(3 xx 10^(-3)) = 2.1 xx 10^(-9) J` |
|
| 104. |
Two charged particles having charges `1 and -1 mu C` and of mass `50 g` each are held at rest while their separation is `2 m`. Now the charges are released. Find the speed of the particles when their separation is `1 m`.A. `(1)/(5) ms^-1`B. `(3)/(5) ms^-1`C. `(3)/(10) ms^-1`D. `(2)/(7) ms^-1` |
|
Answer» Correct Answer - c Applying conservation of mechanical energy, we get `(-9 xx 10^9 xx 10^-12)/(2) = (-9 xx 10^9 xx 10^-12)/(1) + (1)/(2)(2 m) v^2` or `v = 3//10 ms^-1`. |
|
| 105. |
Three plates `A, B, C` each of area `50 cm^(2)` have separation `3 mm` between `A` and `B mm` between `B` and `C`. The energy stored when the plates area fully charged is A. `1.6 xx 10^(-9) J`B. `2.1 xx 10^(-9) J`C. `5 xx 10^(-9) J`D. `7 xx 10^(-9) J` |
|
Answer» Correct Answer - B There are two capacitors parallel to each other. `:.` Total capacitance `= (2epsilon_(0)A)/(d)` `:.` Energy stored ` = (1)/(2) ((2epsilon_(0)A)/(d)) V^(2)` `= (8.86 xx 10^(-12) xx 50 xx 10^(-4) xx 12^(2))/(3 xx 10^(-3)) = 2.1 xx 10^(-9) J` |
|
| 106. |
When the separation between two charges is increase the electric potential energy of the charges.A. increasesB. decreasesC. remains the same.D. may increase or decrease |
|
Answer» Correct Answer - d It depends whether both charges are of the same or opposite sign. |
|
| 107. |
Four identical charges are placed at the points `(1,0,0), (0,1,0),(-1,0,0)`, and `(0,-1,0)`. Then,A. the potential at the origin is zeroB. the electric field at the origin is not zero.C. the potential at all points on the z - axis, other than the origin, is zeroD. the field at all points on the z - axis, other than the origin acts along the z - axis. |
|
Answer» Correct Answer - d All the charges are placed in the x y plane such that they form a square with origin as its center. So electric field at the origin will be zero, but at other points on z - axis. |
|
| 108. |
`ABC` is a right - angled triangle, where `AB and BC` are `25 cm and 60 cm`, respectively. A metal sphere of `2 cm` radius charged to a potential of `9 xx 10^5 V` is placed at `B` as in (Fig. 3.133). Find the amount of work done in carrying a positive charge of 1 coulomb from C to A. .A. 21 k JB. 42 k JC. 14 k JD. 52 k J |
|
Answer» Correct Answer - b Find the potential at `A and C` due to charge at `B`, then the required work done is `W = q(V_A - V_c)`. |
|
| 109. |
Charges`-q , Q, and -q` are placed at an equal distance on a straight liner. If the total potential energy of the system of three charges is zero, then find the ratio `Q//q`. .A. `1//2`B. `1//4`C. `2//3`D. `3//4` |
|
Answer» Correct Answer - b `U = -(k q Q)/( r) - (k qQ)/r + (k q^2)/(2 r) = 0` or `Q //q = 1//4`. |
|
| 110. |
Four charge particles each having charge Q=1 C are fixed at the corners of the base(A,B,C, and D) of a square pyramid with slant length `a(AP=BP=PC=a=sqrt(2)m)`, a charge -Q is fixed at point P. A dipole with dipole moment p=1 Cm is placed at the center of the bases and perpendicular to its plane as shown in fig. Force on the dipole due to the charge particles is `(square)/(4 pi epsilon_(0))N`. |
|
Answer» Correct Answer - i. `F_("dipole") = (3 sqrt(2) Q q)/(pi in. a^2)` (upward) ii. `U - (Q^2)/(2 sqrt(2) pi epsilon . a) - (p Q)/(2 pi epsilon . a^2)`. a. Charges at A , B, C, and D are placed at an equilateral position of dipole. Hence, the force on each of them due to dipole is `F_1 = (Q p)/(4 pi epsilon_0 (a// sqrt(2))^3)` This force is downward on charges. Henc, force due to these charges on dipole is `4F_1` (upward). Force on dipole due to charge at P is `F_2 = (2 p Q)/(4 pi epsilon_0 (a //sqrt(2))^3)` (upward) Net force on dipole is `F = 4F_1 + F_2 = (3 sqrt (2 Q p))/(pi epsilon_0 a^3)` (upward) b. `PE` of the system is `U = (10 "pairs of charged particles") + (5 "pairs of dipole and charged particles")` As potential energy of the dipole with four charges at `A, B, C, and D` will be zero, `U = 4 [(1)/(4 p epsilon_0) (Q^2)/A] + 2 (1)/(4 pi epsilon_0) (Q^2)/(sqrt(2 a)) - 4(1)/(4 pi epsilon_0) (Q^2)/A` `- (1)/(4 pi epsilon_0) (p Q)/((a// sqrt(2))^2)` `U = (Q^2)/(2 sqrt(2) pi epsilon_0 a)- (p Q)/(2 pi epsilon_0 a^2)`. |
|
| 111. |
The potential at a point due to an electric dipole will be maximum and minimum when the angles between the axis of the dipole and the line joining the point to the dipole are respectivelyA. `90^(@) and 180^(@)`B. `0^(@) and 90^(@)`C. `90^(@) and 0^(@)`D. `0^(@) and 180^(@)` |
|
Answer» Correct Answer - D `V = (p cos theta)/(r^(2))`. If `theta = 0^(@)` then `V_(a) = max`. If `theta = 180^(@)` then `V_(e) = min`. |
|
| 112. |
There is `10` units of charge at the centre of a circle of radius `10 m`. The work done in moving `1` unit of charge around the circle once isA. zeroB. `10` unitsC. `100` unitsD. `1` unit |
|
Answer» Correct Answer - A The work done in moving a charge on equipotential surafce is zero. |
|
| 113. |
On increasing the plate separation of a charged condenser, the energyA. increasesB. decreasesC. remains unchangedD. becomes zero |
|
Answer» Correct Answer - A Energy `U = (1)/(2)(Q^(2))/(C )` for a charged capacitor charge `Q` is constant and with the increase in separation `C` will decrease `(C prop (1)/(d))`, So overall `U` will increase. |
|
| 114. |
In an hydrogen atom, the electron revolves around the nucles in an orbit of radius `0.53 xx 10^(-10) m`. Then the electrical potential produced by the nucleus at the position of the electron isA. `-13.6 V`B. `-27.2 V`C. `27.2 V`D. `13.6 V` |
|
Answer» Correct Answer - C `V = 9 xx 10^(9) xx (Q)/(r ) = 9 xx 10^(9) xx ((+1.6 xx 10^(-19)))/(0.53 xx 10^(-10)) = 27.2` |
|
| 115. |
Four identical capacitors are connected as shown in diagram. When a battery of `6 V` is connected between `A` and `B`, the charges stored is found to be `1.5 muC`. The value of `C_(1)` is A. `2.5 muF`B. `15 muF`C. `1.5 muF`D. `0.1 muF` |
|
Answer» Correct Answer - D The capacitance across `A` and `B` `= (C_(1))/(2) + C_(1) + C_(1) = (5)/(2)C_(1)` As `Q = CV`, `1.5 muC = (5)/(2)C_(1) xx 6` `rArrC_(1) = (1.5)/(15) xx 10^(-6) = 0.1 xx 10^(-6) F = 0.1muF` |
|
| 116. |
The energy stored in the condenser isA. `QV`B. `(1)/(2) QV`C. `(1)/(2) C`D. `(1)/(2)(Q)/(C )` |
|
Answer» Correct Answer - B The energy stored `= (1)/(2) QV` |
|
| 117. |
Consider a spherical surface of radius 4 m cenred at the origin. Point charges +q and - 2q are fixed at points A( 2 m, 0,0) and B( 8 m, 0, 0), respectively. Show that every point on the shperical surface is at zero potential. |
|
Answer» If `P (x , y,z)` is any point on the sphere, `x^2 + y^2 + z^2 = (4)^2 = 16` Also `PA = sqrt ((x - 2)^2 + y^2 + z^2)` =`sqrt((x^2 + y^2 + z^2) + 4 - 4 x ) = sqrt(20 - 4 x)` And `PB = sqrt ((x - 8)^2 + y^2 + z^2)` =`sqrt((x ^2 + y^2 + z^2)+ 64 - 16 x)` =`sqrt(80 - 16 x) = 2 sqrt(20 - 4 x)` Thus `V_P = k_e[q/(PA) - (2 q)/(PB)]= k_e [q/(sqrt (20 - 4 x))- (2 q)/(2 sqrt (20 - 4 x))] = 0`. |
|
| 118. |
The potential at any point is given by `V = x(y^2 - 4 x^2)`. Calculate the cartesian components of the electric field at the point. |
|
Answer» Here `V = x(y^2 - 4 x^2)`. So `E_x = -(del V)/(del x) = -(del)/(del x) [xy^2 - 4x^3]` =`- [y^2 - 12x^2]` or `E_x = [12 x ^2 - y^2]` Similarly, `E_y = -(del V)/(del y) = -(del)/(del y) [x (y^2 - 4 x^2)]` or =`- (del)/(del y) [xy^2 - 4 x^3] = -2x y` `E_z = -0` because V does not depend upon the z- coordinate. |
|
| 119. |
(Figure 3.27) shows equipotential surfaces. What is the direction of electric field `vec E` at P and R? . |
| Answer» Electric field lines are perpendicular to the equipotential surfaces and point in the direction of decreasing potential. At P, electric field `vec E` is to the left and at R, `vec E` is upward. | |
| 120. |
Three equipotential surfaces are shown in (Fig. 3.28) A. Draw the corresponding field lines and estimate the field strength at a point A where the distance between the surfaces is 4 cm. a. |
|
Answer» The field lines are perpendicular to the equipotential surfaces as shown in (Fig. 3.28)(b). In the vicinity of a, the surfaces are nearly flat, and so, in order to estimate the field strength, it is a reasonable approximation to use `V_1 - V_b ~~ E. d` or `E ~~ (V_a - V_b)/d = (8 V - 6 V)/(4 xx 10^-2m) = 50 Vm^-1`. |
|
| 121. |
Consider two points `1 and 2` in a region outside a charged sphere. Two points are not very far away from the sphere. If `E and V` respresent the electric fileld vector and the electric potential. Which of the following is not possible ?A. `|vec(E)_(1)|=|vec(E)_(2)|,V_(1)=V_(2)`B. `vec(E)_(1)nevec(E)_(2),V_(1)neV_(2)`C. `vec(E)_(1)nevec(E)_(2),V_(1)=V_(2)`D. `|vec(E)_(1)|=|vec(E)_(1)|,V_(1) ne V_(2)` |
|
Answer» Correct Answer - D |
|
| 122. |
Two capacitors of `3 pF` and `6pF` are connected in series and a potential difference of `5000 V` is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates isA. `2250 V`B. `2222 V`C. `2.25 xx 10^(6)V`D. `1.1 xx 10^(6 V` |
|
Answer» Correct Answer - B `(1)/(C ) = (1)/(3) + (1)/(6) rArr C = 2pE` Total charge `= 2 xx 10^(-12) xx 5000 = 10^(-8) C` The new potential when the capacitors are connected in parallel is `V = (2 xx 10^(-8))/((3 + 6) xx 10^(-12)) = 2222 V` |
|
| 123. |
The variation of potential with distance R from the fixed point is shown in (Fig. 3.125). . The electric field at `R = 5 m` is.A. `2.5 Vm^-1`B. `-2.5 Vm^-1`C. `0.4 Vm^-1`D. `-0.4 Vm^-1` |
|
Answer» Correct Answer - a `E = - (d V)/( d r)` = negative of the slope of `V - r` garph. |
|
| 124. |
Two plates are `2cm` apart, a potential difference of `10` volt is applied between them, the electric field between the plates isA. `20 N//C`B. `500 N//C`C. `5 N//C`D. `250 N//C` |
|
Answer» Correct Answer - B `E = (V)/(d) = (10)/(2 xx 10^(-2)) = 500 N//C` |
|
| 125. |
Which of the following is true for the figure showing electric lines of force ? (E is electrical field, and V is potential). .A. `E_A gt E_B`B. `E_B gt E_A`C. `V_A gt V_B`D. `V_B gt V_A` |
|
Answer» Correct Answer - a.,d. From electric field line pattern, it is clear `|E_A|gt|E_B|` and electric potential decreases in the direction of field, hence `V_A lt V_B`. |
|
| 126. |
The electrical potential function for an electrical field directed parallel to the x-axis is shown in the given graph. Draw the graph of electric field strength.A. `-2 le x le 0, 4 le x le 8`, and `0 le x le 2`B. `-2 le x le 0` and `0 le x le 2`C. `-2 le x le 0, 2 le x le 4`, and `le x le 8`D. `0 le x le 2` and `4 le x le 8` |
|
Answer» Correct Answer - C `E_(r)=-(DeltaV)/(Deltar), E_(x)=-(DeltaV)/(Deltax)` For `-2 le x le 0, E_(x)=-((10-0))/([0-(-2)])=-5 NC^(-1)` For `0 le x le 2, E_(x)=0` For `2 le x le 4, E_(x)=-([20-10])/([4-2])=-5 NC^(-1)` For `4 le x le 8, E_(x)=-([0-20])/([8-4])=5 NC^(-1)` |
|
| 127. |
The electrical potential function for an electrical field directed parallel to the x - axis is shown in (Fig. 3.152). . The magnitude of the electric field in the x - direction in the interval `4 le x le 8` isA. `2.5 NC^-1`B. `5 NC^-1`C. `-2.5 NC^-1`D. `-5 NC^-1` |
|
Answer» Correct Answer - B `E_(r)=-(DeltaV)/(Deltar), E_(x)=-(DeltaV)/(Deltax)` For `-2 le x le 0, E_(x)=-((10-0))/([0-(-2)])=-5 NC^(-1)` For `0 le x le 2, E_(x)=0` For `2 le x le 4, E_(x)=-([20-10])/([4-2])=-5 NC^(-1)` For `4 le x le 8, E_(x)=-([0-20])/([8-4])=5 NC^(-1)` |
|
| 128. |
A uniform electric field is present in the positive x - direction. If the intensity of the field is `5 NC^-1` then find the potential difference `(V_B - V_A)` between two points A (0 m, 2 m) and B (5 m, 3 m). |
|
Answer» we know potential difference between points A and B in uniform electric field is `V_B - V_A = - vec(E) . vecA B`. Here we are given `vec(E) = 5 hat(i) (Vm^-1)` and `vec A B = (5 - 0) hat(i) + (3 - 2) hat(j) (m)` Hence, `V_B - V_A = -(5 hat(i)).(5 hat(i) + hat(j))= - 25 V`. |
|
| 129. |
A uniform field of magnitude `vec E = 2000 NC^-1` is directed `theta = 37^(@)` below the horizontal [fig. 3.11]. . (i) Find the potential difference between P and R. (ii) If we define the reference level of potential so that potential at R is `V_R = 500 V`, what is the potential at P? |
|
Answer» (i) Here `E = 2000 NC^-1, theta = 37^(@), V_R = 500 V` `Delta l = PR = 5 cm = 5 xx 10^-2 m` `V_R - V_P = vec E . Delta vec l = - E d l cos theta` =`-(2000 NC^-1)(5 xx 10^-2 m) cos 37^(@)` `V_R - V_P = - 80 V` Thus, `V_P - V_R = vec E . Delta vec l = 80 V` (ii) As `V_P- V_R = 80 V and V_R = 500 V`, `V_P = 500 V + 80 V = 580 V`. |
|
| 130. |
If potential (in volts) in a region is expressed as `V (x, y, z) = 6xy - y + 2yz`, the electric field (in `N//C)` at point `(1, 1, 0)` isA. `-(6 hati + 9hatj + hatk)`B. `-(3hati + 5hatj + 3hatk)`C. `-(6 hati + 5 hatj + 2hatk)`D. `-(2hati + 3hatj + hatk)` |
|
Answer» Correct Answer - C `V (x, y, z) = 6xy - y + 2yz` `E_(X) = - (delV)/(delx) =- 6y = - 6` `E_(y) = - (delV)/(dely) = - 6x + 1+ 2z = - 5` `E_(z) = - (delV)/(delz) = - 2y = - 2` `vecE = E_(x) hati + E_(y) hatj + E_(z) hatk` `vecE = - (6 hati + 5 hatj + 2 hatk)` |
|
| 131. |
An ail drop having charge `2e` is kept stationary between two parallel horizontal plates `2.0 cm` apart when a potential difference of `12000` volts is applied between them. If the density of oil is `900 kg//m^(3)`, the radius of the drop will beA. `2.0 xx 10^(-6) m`B. `1.7 xx 10^(-6) m`C. `1.4xx 10^(-6) m`D. `1.1 xx 10^(-6) m` |
|
Answer» Correct Answer - B In equilibrium `QE = mg` `rArrQ. (V)/(d) = mg = ((4)/(3)pi r^(3) rho) g` `rArr 2 xx 1.6 xx 10^(-19) xx (12000)/(2 xx 10^(-2)) = (4)/(3)pi r^(3) xx 900 xx 10` `rArr r = 1.7 xx 10^(-6)m` |
|
| 132. |
In a certain region, electric field `E` exists along the x - axis which is uniform. Given `B = 2 sqrt(3) m` and `BC = 4 m`. Points A, B, and C are in `x y` plane. . A charged particle `q` is moved from A to C as shown in path `1`. What is the work done by electric field in this process ?A. 2 q EB. 5 q EC. q ED. 4 q e |
|
Answer» Correct Answer - B `W_(e 1) = q (V_A - V_C)` =`q(V_A - V_B) + (V_B - V_C) = q (3 E + 2 E) = 5 q E`. |
|
| 133. |
An electron enters in high potential region `V_(2)` from lower potential region `V_(1)` then its velocityA. will increasesB. will change in direction but not in magnitudeC. bo change in direction of filedD. no change in direction perpendicular to field |
|
Answer» Correct Answer - A Electron is moving in opposite direction of field so field will produce an accelerating effect on electron. |
|
| 134. |
Assertion: Circuit containing capacitors should be handled cautiously even when there is no current. Reason: The capacitors are very delicate and so quickly break down.A. If both assertion and reason are true and reson is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
|
Answer» Correct Answer - C A charged capacitor, after removing the battery, does not discharge itself. If this capacitor is touched by someone, he may feel shock due to large charge still present on the capacitor. Hence it should be handled cautiously otherwise this may cause a servere shock. |
|
| 135. |
A `20F` capacitor is charged to `5V` and isolated. It is then connected in parallel with an uncharged `30 F` capacitor. The decrease in the energy of the system will beA. `25 J`B. `200 J`C. `125 J`D. `150 J` |
|
Answer» Correct Answer - D Charge stored on `20 F` capacitor `= 20 xx 5 = 100 C` When `20F` capacitor is connected to `30F` capacitor, `V = (q)/(C_(1) + C_(2)) = (100)/(20 + 30) = 2V` Energy of the system will be `E_(f) = (1)/(2) xx (30 + 20) xx (2)^(2)` `= 100 J` Energy before isolation `E_(i) = (1)/(2) xx 20 xx (5)^(2) = 250 J` Therefore, decrease in energy `= 250 - 100 = 150 J` |
|
| 136. |
Assertion: An isolated system consist of two particles of equal masses `m = 10 gm` and charged `q_(1) =+1muC` and `q_(2) =-1muC` as shown in figure. The initial separation of both charges is `l = 1m`. Both the charges are given initial velocities `v_(1) = 1m//s` and `v_(2) = 2m//s` towards right. then the maximum separation between the charges is infinity. Reason: The total energy (Kinetic energy + electrostatic potential energy) of given two particle system is positive and initial velocity of separation is positive.A. If both assertion and reason are true and reson is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
|
Answer» Correct Answer - A `K.E.` of system of two particles `(1)/(2)mv_(1)^(2) + (1)/(2)mv_(2)^(2) = (1)/(2) 10 xx 10^(-3) xx 1^(2) + (1)/(2) 10 xx 10^(-3) xx 2^(2) = 2.5 xx 10^(-2) J` Electrostatic potential energy of system of two particles is `= (1)/(4pi epsilon_(0)) (q_(1)q_(2))/(l) = 9 xx 10^(9) xx (1 xx 10^(-6) xx (-1) xx 10^(-6))/(1)` `=-9 xx 10^(-3) J` `:.` Total energy of system `= 2.5 xx 10^(-2) - 9 xx 10^(-3)` positive Since the total energy of system is positive, the system is unbounded. therefore maximum distance between particles is infinity. |
|
| 137. |
Both the ring and the conducting sphere are given the same charge `Q`. Determine the potential of the sphere. Assume that the centre of the sphere lies on the axis of th ring. Is it necessary that the charge on the ring be uniformly distributed to answer the above question ? |
|
Answer» Correct Answer - `(kQ)/(Rsqrt(10))+(kQ)/(R),No` |
|
| 138. |
A charge +q is carried from point `A( r, 135^(@))` to point `B(r, 45^(@))` following a path, which is a quadrant of circle of radius r ( Fig. 3.97). If the dipole moment is P, find the work done by the external agent (assume short dipole). .A. `0`B. `(1)/(4pi epsilon_(0))(qp)/(r^(2))`C. `(1)/(4pi epsilon_(0))(sqrt(2)qp)/(r^(2))`D. `(1)/(4pi epsilon_(0))(qp)/(r )` |
|
Answer» Correct Answer - C `V_(A) = (kp cos 135^(@))/(r^(2))` `V_(B) = (kp cos 45^(@))/(r^(2))` work done `= q (V_(B) - V_(A))` `= (1)/(4pi epsilon_(0)) (sqrt(2) pq)/(r^(2))` |
|
| 139. |
The charge flowing through the cell on cloing the key `k` is equal to : A. `(CV)/(4)`B. `4CV`C. `(4)/(3)CV`D. `(3)/(4)CV` |
|
Answer» Correct Answer - A When switch is open, the charge given to equivalent capacitor. `q_(i) = ((3C xx C)/(3C + C)).V = (3)/(4)CV` When switch is closed, `q_(f) = CV` `Deltaq = q_(f) - q_(i) = CV - (3)/(4)CV = (1)/(4)CV` Hence charge passed through switch will be `(1)/(4)CV`. |
|
| 140. |
In the figure a capacitor id filled with dielectric. The resultant capacitance is A. `(2epsilon_(0)A)/(d) [(1)/(k_(1)) + (1)/(k_(2)) + (1)/(k_(3))]`B. `(epsilon_(0)A)/(d) [(1)/(k_(1)) + (1)/(k_(2)) + (1)/(k_(3))]`C. `(2epsilon_(0)A)/(d) [k_(1) + k_(2) + k_(3)]`D. None of these |
|
Answer» Correct Answer - D `C_(1) = (K_(1)epsilon_(0)(A)/(2))/(((d)/(2))) = (K_(1)epsilon_(0)A)/(d)` `C_(2) = (K_(2)epsilon_(0)((A)/(2)))/(((d)/(2))) = (K_(2)epsilon_(0)A)/(d)` and `C_(3) = (K_(3)epsilon_(0)A)/(2d) = (K_(3)epsilon_(0)A)/(2d)` Now, `C_(eq) = C_(3) + (C_(1)C_(2))/(C_(1) + C_(2)) = ((K_(3))/(2) + (K_(1)K_(2))/(K_(1) + K_(2))).(epsilon_(0)A)/(d)` |
|
| 141. |
It is possible to obtain a large potential difference by charging a number of capacitors in parallel and then disconnecting them to connect them again in series. What maximum potetnial difference can be obtained using `10` capacitors each of `5 muF` and a charging source of `400 V` ?A. `4000 V`B. `2000V`C. `400 V`D. `1000 V` |
|
Answer» Correct Answer - A In parallel , potential of each capacitor will be `400 V`. In series all these potentials will be added to give `4000 V`. |
|
| 142. |
A negative charge is moved by an external agent in the direction of electric field. Then .A. Potential energy of the charge increases.B. Potential energy of the charge decreases.C. Positive work is done by the electric fieldD. negative work is done by the electric field |
|
Answer» Correct Answer - a.,d. When a negative charge moves from high potential to low potential, its potential energy increases. `W_(e 1) = q(V_1 - V_2)` As `V_1 gt V_2` and `q` is negative, `W_(e 1)` is negative. |
|
| 143. |
At a distance of `5 cm and 10 cm` outward from the surface of a uniformly charged solid sphere, the potentials are `100 V and 75 V`, repectively. Then.A. Potential at its surface is `150 V`B. the charge on the sphere is `(5//3) xx 10^-10 C`C. the electric field on the surface is `1500 Vm^-1`D. the electric potential at its center is `0 V` |
|
Answer» Correct Answer - a.,c. `V = (1)/(4 pi epsilon_0) q/((R + d))` `R` is radius, and `d` is distance from surface `100 = K q/((R + 5) xx 10^-2))`…(i) `75 = K q/((R + 10) xx 10^-2))` ….(ii) From (i) and (ii). `R = 10 cm` and `Q = (50)/(3) xx 10^-10 C` Potential at surface is `V_0 = (k Q)/(R) = (9 xx 10^9 xx 50)/(10 xx 10^-2 xx 3) xx 10^-10 = 150 V` Electric field on surface is `E_0 = (k Q)/(R^2) = 1500 Vm^-1` Potential at the center is `V_C = (3)/(2) V_0 = 225 V`. |
|
| 144. |
A uniform electric field of magnitude `325 Vm^-1` is directed in the negative y - direction in (fig. 3.12). Theb coordinates of point A are `(-0.2 m , -0.3 m)` and those of point B are `(0.4 m, 0.5 m)`. Calculate the potential difference `V_B - V_A` along the path shown in the figure. . |
|
Answer» We can define the potential difference between the points A and B. `V_B - V_A = - undersetAoversetB(int) vec E. vec d s` = `- undersetAoversetB(int) (E_x hat i + E _y hat j + E _z hat k).(d x hat i + d y hat j + d z hat k)` = `- [ E_x underset(x_ A)overset(x_B)(int) dx + E_y underset(y_ A)overset(y_B)(int) d y + E_z underset(z_ A)overset(z_B)(int)d z]` =`-[E_x (x_B - x_A) + E_y (y_B - y_A) + E_z(z_B - z_A)]` =`-[( -325)^5 {(0.5) - (-0.3)^3} ] = 325 xx 0.8 = 260 V` `V_B - V_A = + 260 V`. |
|
| 145. |
Uniform electric field of magnitude `100 Vm^-1` in space is directed along the line `y = 3 + x`. Find the potential difference between points `A (3,1) and B (1,3)`. |
|
Answer» Equation of the line `y = 3 + x`. The slope of line `tan theta = 1 or theta = 45^(@)` we can express electric field in vector form `vec E = 100 cos theta hat i + 100 sin theta hat j (V m^-1)` Electric field `vec E = |vec E | (cos theta hat i + sin theta hat j)` `vec E = (100)/(sqrt(2)) hat i + (100)/(sqrt(2)) hat j (V m^-1)` `Delta vec r = vec (AB) = vec r_B - vec r _A = (hat i + 3 hat j)-(3 hat i + hat j)= -2 hat i + 2 hat j` `Delta V = - vec E . Delta vec r = - ((100)/(sqrt(2)) hat i + (100)/(sqrt(2)) hat j). Delta vec r` = `- ((100)/(sqrt(2)) hat i + (100)/(sqrt(2)) hat j) . (-2 hat i + 2 hat j)` =`-100 sqrt(2) + 100 sqrt(2)` `V_A - V_B = 0`. |
|
| 146. |
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 remain in equilibriumB. It can move inside `S`, but it cannot cross `S`C. It must cross `S` at some timeD. It may move, but will ultimately return to its starting point |
|
Answer» Correct Answer - C A free positive charge move from higher (positive) potential to lower (negative) potential. Hence, it must cross `S` at some time. |
|
| 147. |
An electric field is given by `E_(x) = - 2x^(3) kN//C`. The potetnial of the point `(1, -2)`, if potential of the point `(2, 4)` is taken as zero, isA. `-7.5 xx 10^(3) V`B. `7.5 xx 10^(3) V`C. `-15 xx 10^(3) V`D. `15 xx 10^(3) V` |
|
Answer» Correct Answer - A `dV = - vecE.vecdr = - (-2x^(3)hati).(dxhati + dyhatj + dz hatk) = 2x^(3) dx` `rArr underset(0)overset(v)int dV = underset(2)overset(1)int (2x^(3)) xx 10^(3) dx` `V = - 7.5 xx 10^(3) V` |
|
| 148. |
Two spherical conductors each of capacity `C` are charged to potetnial `V` and `-V`. These are then conneted by means of a fine wire. The loss of energy will beA. ZeroB. `(1)/(2)CV^(2)`C. `CV^(2)`D. `2CV^(2)` |
|
Answer» Correct Answer - C `DeltaV = (1)/(2)(C xx C)/((C + C)) |V - (-V)|^(2) = CV^(2)` |
|
| 149. |
find the work done by an external agent in slowly shifting a charge `q = 1 mu C` in the electric field ` vec E = 10^3 hat i Vm^-1` from the point `P(1,2) "to" Q(3, 4)`. |
|
Answer» The work done by the external agent against the electric field is `W_(ext) = -q underset(1) overset(2) (int) vec E . vec d l = -qint (E hat i) .(dx hat i + dy hat j) ` = `q underset(x_1) overset(x_2)(int) E dx = q E(x_2 - x_1) = (10^-6)(10^3)(3- 1)= 2 xx 10^-3 J`. |
|
| 150. |
The charge given to a hollow sphere of radius `10 cm` is `3.2 xx 10^(-19)` coulomb. At a distance of `4 cm` from its centre, the electric potetnial will beA. `28.8 xx 10^(-9)` voltsB. `288` voltsC. `2.88` voltsD. Zero |
|
Answer» Correct Answer - A Potential is to be determined at a distance of `4 cm` from centre of sphere i.e., inside the sphere. `V = (kq)/(r ) = (9 xx 10^(9) xx 3.2 xx 10^(-19))/(10 xx 10^(-2)) = 28.8 xx 10^(-9)` volts |
|