Sphere C carries no net charge. Case 2: at a point inside the spherical shell soon. p = q × 2a = 5 × 10 -6 × 10-3 = 5 × 10-9 Cm. The two fields at P are in opposite directions. Charge are again shared 8. Point is outside the spherical shell of radius ‘a’ but inside the spherical Shell of radius ‘b’.
For derivation see sol.
Why do the electric field lines never cross each other? A charge +Q is uniformly distributed within a sphere of radius R. Find the electric field, distant r from the center of the sphere where: (1) r < R and (2) r > R. (CBSEAI 2016C) Spheres A and B are then brought in contact and then separated . 12.
TWO identical metallic spheres A and B having charges 4Q and -m kept a certain distance apart.
Answer: Using Gauss law, derive an expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius Rand charge density C/m2. What is the total electric flux leaving the surface of the sphere?
Question 21. over here on EduRev! Why do the electric field lines not form closed loops? One is the Law of Conservation of Charge - when two spheres touch the total charge on the two afterwards equals the total charge they had before. Found inside – Page P-68Two small similar metal spheres A and B having charges 4q and – 4q, when placed at a certain distance apart, exert an electric force F on each other. When another identical uncharged sphere C, first touched with Athen with B and then ... If the dipole is rotated from θ1 = \(\frac{\pi}{2}\) to θ2 =θ, then, U = – pE(cos θ – cos \(\frac{\pi}{2}\) ) Question 28. Answer:
Or Hence electric flux through the square (one face of the cube) (NCERT Exemplar) An infinite line charge produces a field of 9 × 10 4 N/C at a distance of 2 cm.
Mathematically the electric flux passing through an area \(\vec{dS}\) is given by
Or straight wire of linear charge density λ C/m.
Answer: Given: Repulsive force of magnitude 6 × 10 −3 N. Charge on the first sphere, q 1 = 2 × 10 −7 C. Charge on the second sphere, q 2 = 3 × 10 −7 C. Distance between the spheres, r = 30 cm = 0.3 m \(\oint_{s} \vec{E} \cdot \vec{d} S=\frac{0}{\varepsilon_{0}}\) [∵ No charge exists inside the spherical shell] Answer:
Therefore, we have 42. Hence by Gauss’s law, there is no net charge enclosed by the closed surface. Derive an expression for the torque acting on an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\). Answer: Two identical plane metallic surfaces A and B are kept parallel to each other in the air, separated by a distance of 1 cm as shown in the figure. when the electric field everywhere on the surface be zero then the net charge inside it must be zero.
Use Gauss’s law to derive the expression for the electric field (\(\vec{E}\) ) due to a straight uniformly charged infinite line of charge λ Cm-1.
Two similarly and equally charged identical metal spheres A and B repel each other with a force of 2x10-5 N. A third identical uncharged sphere C is touched with A and then placed at the midpoint between A and B. Q.13 Two charged metallic sphere are joined by a very thin metal wire. (CBSE 2019C)
(b) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2 > r1).
The electric flux through a given surface is defined as the dot product of the electric field and area vector over that surface.
Sketch electric field lines originated from the point on to the surface of the plate. Answer:
Therefore, the net force acting on the dipole is Sphere C is first touched to A, then to B, and then removed. Answer: Determine the electric field (i) between the sheets, and (ii) outside the sheets. Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? Of the resultant electric field at the vertex A due to these two charges. (i) Define electric flux. The charge distribution is spherically symmetric. Or.
Since all the three spheres are identical, i.e. The electric field lines are as shown. Since the electric field has only an x component, for faces perpendicular to the x-direction, the angle between E and ΔS is ± π/2. (a) Spherical Shell Three cases arise, Case 1: at a point outside the spherical shell Q:-A polythene piece rubbed with wool is found to have a negative charge of 3 × 10 −7 C. On inner surface
Let us suppose that the dipole is placed in a vacuum. Question 5. (CBSE Delhi 2011) A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B, then spheres A and B are brought in contact and then separated. No Marketing Blurb
(4), (b) An infinitely large thin plane sheet has a uniform surface charge density +σ.
Answer: Question 8.
Two metallic spheres A and B having charges +4q and -4q respectively when placed at certain distance apart exert an electric force of 'F' on each other When another identical uncharged spenere 'C', first touched with A and then touched with B and then removed to infinity What will - Physics - Electric Charges And Fields Each sphere has the same charge q 1 = q 2 = q. Q:-What is the force between two small charged spheres having charges of 2 x 10-7 C and 3 x 10-7 C placed 30 cm apart in air?
(CBSE Delhi 2013) Answer: Sphere C is then touched to sphere A and seperated from it. P.E. Answer: And θ = 90° – 30° = 60°, The electric flux linked with the square sheet Answer: Φ = q/ 4ε0 The plates have surface charge densities of opposite signs and of magnitude 2.0 x 10-12 C/m2. b. Two small similar spheres A and B having charges 4q and -4q, when placed at a certain (a) By Gauss theorem \(\phi \vec{E} \cdot \overrightarrow{d S}=\frac{q}{\varepsilon_{0}}\). Three identical metallic uncharged spheres A, B and C each of radius a, are kept at the corners of an equilateral triangle of side d(d>>a) as shown in Fig. The same amount of charge is removed from each sphere. The radius of each sphere is very small compared to the distance between them, so that they may be considered as point charges.
(ii) For a < x < b (a) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) The charge will be shared by two cubes if it has to be enclosed. A new edition of a classic textbook, introducing students to electricity and magnetism, featuring SI units and additional examples and problems. units. What is the force between two small charged spheres having charges of 2 ×10-7 C and 3 × 10 -7 C placed 30 cm apart in air?. = – Eds1 + Eds2 (0) + Eds cos 0° It is measured in N m2 C-1, (ii) Consider a uniform electric field \(\vec{E}\) = 5 × 103î NC-1. Consider a uniformly charged ring of radius ‘a’. Force due to charge Q placed at point A is Work done, Question 1.
Question 11.
Answer: A positive point charge (+q) is kept in the vicinity of an uncharged conduction plate. A third distance apart. Point charges +49, -q and +4q are kept on the X- axis at point x = 0, x = a and x = 2a respectively. Find the position and nature of Q. Found inside – Page 247Can two equipotential surfaces intersect? Two identical metallic spheres A and B carry charges of +Q and –Q, respectively. The force between them is F newtons when they are separated by a distance d in air. The spheres are allowed to ...
Draw a plot showing the variation of an electric field (E) with distance, 15.
Two large parallel plane sheets have uniform charge densities +σ and -σ. A charge of 17.7 × 10-4 C is distributed uniformly over a large sheet of area 200 m2. (a) Derive the expression for the electric field at a point on the equatorial line of an electric dipole.
A point charge of +10 μC is placed at a distance of 20 cm from another identical . 6. Question 8. Explain. is maximum when cos θ = – 1, i.e.
If third identical uncharged sphere is first placed in contact with each other and then with sphere B, then the spheres A and B are brought in contact with each other and separated. Charge on inner surface – Q. 21-21a). Therefore. 14. For position P1 dipole moment and electric field are parallel.
Four identical metal spheres have charges of q A =-8.0 µC, q B =-2.0 µC, q C = + 5.0 µC, and q D = + 12.0 µC. The diagram is as shown.
It is measured in Nm2C--1.
(CBSE Delhi 2018) Thus there will be two directions of the electric field at that point which is not possible. ∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p. (b) Two identical point charges, q each, are kept 2 m apart in the air. Φ = q/ 2ε0 An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N/C (figure a) If the radius of the larger sphere is twice that of the smaller one, the electric field near the larger sphere is . therefore \(\frac{F}{3 F}=\frac{r^{2}}{d^{2}}=\frac{1}{3} \Rightarrow r=\frac{d}{\sqrt{3}}\), Question 22. Question 26.
Deduce the necessary expression. (ii) a ≤ x < b What will be the total flux through the faces of the cube (figure) with the side of length ‘a’ if a charge q is placed at
A third identical uncharged sphere C is first placed in contact with spheres A and then with B, then spheres A and B are brought in contact and then separated. If it experiences a torque of 8\(\sqrt{3}\) Nm, calculate the (i) magnitude of the charge on the dipole and (ii) potential energy of the dipole.
V1 = Q .
(CBSE 2019C)
FA = k\(\frac{Q^{2}}{(a \sqrt{2})^{2}}\) =k\(\frac{Q^{2}}{2 a^{2}}\) along AC, Force due to the charge q placed at D The magnitude of etectñc field at P due to the segment is Let the new distance be ‘d’, since F ∝\(\frac{1}{r^{2}}\) , If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence. (CBSE Al 2013) How q = 8N\(\sqrt{3}\) × 2 / 2 × 10-2 × 105 × \(\sqrt{3}\) = 8 × 10-3 C, (ii) Using U = -pE cos θ, we have Hence, the initial force Fi = k*Q*3Q/r^2 and the final force Ff = k*2Q*2Q/r^2. From the diagram we have r = \(\sqrt{x^{2}+a^{2}}\) and cos α = x/r, therefore we have Find the charge on the spheres A and B.
Calculate the net outward flux through the cylinder. Calculate the net electric force on C. 11. There are two main ideas here. Therefore, the net field within the dielectric decreases to \(\vec{E}_{0}\) – \(\vec{E}\), Question 5.
q1 is negative and q2 is positive. (i) It is defined as the number of electric field lines crossing a unit area perpendicular to the given area. Now the force of repulsion between them is 0.025 N. Calculate the final charge on each of them.
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\), (b) Discuss the orientation of the dipole in (a) stable, (b) unstable equilibrium in a uniform electric field.
(CBSE Delhi 2018C) Thus, a hollow charged conductor is equivalent to a charged spherical shell. If the point of observation is far away, i.e. The mutual forces between two charges do not get affected by the presence of other charges. two identical metallic ball spherical shells A and B charges 4Q and -10c are kept at a certain distance apart, a third uncharged sphere C is first placed in contact with sphere A and then with . Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\). The charges begin to leak from both the spheres at a constant rate. Torque on an electric dipole in a uniform electric field, τ = PE sin θ. U = -pE cos θ, the potential energy of an electric dipole.
Answer: Why?
In which orientation, a dipole placed in a uniform electric field is in. A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B. Two identical metallic spheres, having unequal, opposite charges are placed at a distance of 0.90 m in air. Inner surface, As a result of this symmetry, we consider a Gaussian surface in the form of a cylinder with arbitrary radius r and arbitrary Length L. with its ends perpendicular to the wire as shown in the figure. Use Gauss’s law to find the expressions for the electric field at points P1 and P2.
τ = pEsinθ …(i) (a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole. Charge on C after it is in contact with B Found inside – Page 20-14When two identical metallic spheres are brought in contact , the charges on them are equalized due to the flow of free electrons . Thus when an uncharged sphere C is brought in contact with sphere A having a charge + ... respectively as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the sphere S1 change if a medium of dielectric constant ‘er’ is introduced in the space inside S1 in place of air? A uniformly charged long straight wire. Find the charges on the sphere A and B.
Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams. Therefore, the flux is …(2).
Draw a plot showing the variation of an electric field (E) with distance r due to a point charge Q.
This book basically caters to the needs of undergraduates and graduates physics students in the area of classical physics, specially Classical Mechanics and Electricity and Electromagnetism. State the Gauss law of electrostatics. A conducting sphere having a negative charge on it. For r > R Let P be the point of observation on the axial Line where the electric field has to be found. A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B, then spheres A and B are brought in contact and then separated. (a) Two of the spheres are brought together so they touch, and then they are separated.
Using this law, derive an expression for the electric field due to a Initially the spheres are separated by a distance r. The spheres briefly touch each other and move to the initial separation. Q.39 Two metallic spheres A and B having charges +4Q and -10Q are kept a certain distance apart.
(a) State Gauss’s law. 17.
Question 17. Now the uncharged sphere is earthed . (d) D: mid-point of B and C. {NCERJ Exemplar) Point Lies inside both the spherical shells. Answer: x = \(\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})\) to the left of q. Φ = E S cos 60°
Found inside – Page 739Two solid spheres, both of radius 5 cm, carry identical total charges of 2 mC. Sphere A is a good conductor. Sphere B is an insulator, and its charge is distributed uniformly throughout its volume. (i) How do the magnitudes of the ...
(d + x)2 = 3x2, Solving for x we have
Question 4. = 200 × 0.01 × \(\frac { 1 }{ 2 }\) = 1.0 Nm2 C-1. (a) An electric dipole of dipole moment \(\vec{p}\) consists of point charges +q and -q separated by a distance 2a apart. Found inside – Page 7664 identical drops each of capacity 5 μF combine to (a) directed perpendicular to the plane and away from form a big drop. ... Two metallic spheres of radii 1 cm and 3 cm are given charges of (–1 × 10–2 C) and (5 × 10–2 C), ... (b) Two identical metallic spheres A and B having charges +4Q. E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = EB – EA, since θ = 1800 Therefore, the resultant electric field is, Solving we have Two charges of magnitude – 2Q and + Q are located at points (a, 0) and (4a, 0), respectively. Why must the electrostatic field be normal to the surface at every point of a charged conductor?
An uncharged insulated conductor A is brought near a charged insulated conductor B. For stable equilibrium \(\vec{P}\) is along \(\vec{E}\)
Draw the field lines when the charge density of the sphere.
There are two uncharged identical metallic spheres 1 and 2 of radius r separated by a distance d (d > > r) . Draw a graph to show the variation of E with perpendicular distance r from the line of charge (CBSE Delhi 2018)
about the charges ? (iii) For b ≤ x< ∞ Q.24 Two equally charged identical metal spheres A and B repel each other with a force 3 .
Answer: = -.125 μs, Question 11. F 24 = F 2q + F 4q = F 2 cos 55° + F 4 cos 45 . 10. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at vertex A due to these two charges. Question 22. U = – pE When θ = 0°, Question 4. When an electric dipole is placed parallel to a uniform electric field, net force, as well as net torque acting on the dipole, is zero and, thus, the dipole remains in equilibrium. Answer:
E ∝ \(\frac{1}{r}\).
Using this law derive an expression for the electric field due to a long (CBSE Al, Delhi 2018)
(a) Define electric dipole moment. In the figure given below, at which point electric field is maximum?
What will be the surface charge density on the (i) inner surface, and (ii) the outer surface of the shell? what is the physical significance of this limit? Gauss’s law states that the net outward flux through any closed surface is equal to 1 /ε0 times the charge enclosed by the closed surface. Answer:
(a) No two field lines can cross, because at the point of intersection two tangents can be drawn giving two directions of the electric field which is not possible. Point P2 lies inside the metal, therefore the Gaussian surface drawn at P2 does not include a charge, hence the electric field at P2 is zero. What is the new charge on each sphere?
electric field lines and why? Answer: The zero potential points lie on the equatorial line. Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 x 10-2 C and 5 x 10-2 C respectively. Fmed = \(\frac{1}{4 \pi \varepsilon_{0} K} \frac{q_{1} q_{2}}{r^{2}}\) where K is dielectric constant. (ii) Two identical metallic spheres a and b having charges +4Q and -10Q are kept a certain distance apart. PR = \(\sqrt{p^{2}+p^{2}+2 p^{2} \cos 120^{\circ}}\) = p, since cos 120° = -1/2
Φ = E × 4πr² …(1), But by Gauss’s law For unstable equilibrium \(\vec{P}\) is antiparallel to \(\vec{E}\) Find the charge on the spheres A and B. Starting again from the initial conditions what is the charge on each sphere if C is touched to A removed and then touched to B. (i) parallel to the field Three-point charges q, -4q, and 2q are placed at the vertices of an equilateral triangle ABC of side T as shown in the figure. ii. The required graph is
Finally, Ff = 4/3 * Fi = 4/3*48 = 64 µN. where p = q 2a. Therefore, q = 2a3 × 8.854 × 10-12C. Two charges, one +5 µC, and the other -5 µC are placed 1 mm apart.
through a smaLL displacement ‘dr’, dW = \(\vec{F}\) . Define electric field intensity at a point.
Asked by kashish bhatia; Two point charges q1andq2 , of magnitude 10^-8c and-10^8c , respectively, are placed 0.1m apart.
(CBSE Al 2012C) Question bank for NEET. Area of square = 12 × 10-4 m2, The plane of surface area being parallel to YZ plane, hence Point out right or wrong for the following statement. The way the electric field strength (E) of a point charge q weakens with (r) is like the way light intensity weakens as we move away from a light bulb. (i) Find out the ratio of the electric flux through them. E = EA cos θ + EB cos θ 5. The electric field intensity is zero at the point P on the line joining them as shown. 1. The arrows p1 and p2 show the directions of its electric dipole moment in the two cases. Φ = Φ1 – Φ2 = [Ex(atx= 2a) – Ex (atx = a)]a2 Find the charge on the spheres A and B. 10. Question 26.
Consider a spherical shell of radius R. Let q be a charge on the shell. x >> a, then E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{x^{2}}\). Two small similar metal sphere A and B having charge 4 q and -4q when placed at a certain distance apart exert an electric force F on each other when another identical uncharged sphere C first touched with A then with B and then removed to infinity the force of interaction between A and B for the same separation will be (a) F/2 (b) F/8 (c) F/16 (d) F/32? The electric fields due to the two charges placed at B and C are inclined at an angle θ = 120° as shown (CBSE AI 2014) Calculate the charge within the cube, assuming a = 0.1 m. (CBSE Sample Paper 2019)
Deduce an expression for the electric field at a point on the equatorial plane of an electric dipole of length 2, 5. b. EA COS θ and EB cos θ being in the same direction add up as shown in the figure. (CBSEAI 2019) s = \(\frac { 1 }{ 2 }\) at2 [u = 0] (CBSEAI 2011C)
Define the electric line of force and give its two important properties. 1. It is defined as the product of the magnitude of either of the two charges and the distance between them.
Two identical metallic spherical shells A and B having charges +4Q and -10 Q are kept a certain distance apart. (c) A Square plane sheet of side 10 cm is inclined at an angle of 30° with the direction of a uniform electric field of 200 NC“1. Electric Flux is the dot product of the electric field and area vector. Two identical conducting spheres A and B, carry equal charge. Found inside – Page 14Derive an expression for the electric field due to an infinitely long straight wire of linear charge densityλ Cm–1. ... (b) Consider two hollow concentric spheres, S respectively 1 and S2, enclosing charges 2Q and 4Q as shown in the ... (b) B: mid-point of an edge of the cube.
perpendicular to and parallel to AB.
= –pEcos θ they have the same capacity, therefore when uncharged sphere C is placed in contact with A, the total charge is equally shared between them. On bringing uncharged conductor A near a charged conductor B, charges are induced on A as shown in the figure below. (CBSE Sample Paper 2018-2019). Use Gauss’ law to derive the expression for the electric field between two uniformly charged parallel sheets with surface charge densities and -, respectively. Hence there will be a net non-zero force on the dipole in each case. Therefore surface charge density on the inner and the outer shell is on the outer surface is. CD-ROM contains: Demonstration exercises -- Complete solutions -- Problem statements. Answer: (b) The field lines are always perpendicular to the surface of a charged conductor. Answer: \(\vec{dr}\) =q\(\vec{E}\) .\(\vec{dr}\) Two identical metallic spherical shells A and B having charges +4Q and -10Q are kept at 3m apart.
Hence show that for points at a large distance from the ring, it behaves like a point charge. A point charge of +10 μC is placed at a distance of 20 cm from another identical . For the charge 2q to experience zero force we have Deduce the expression for the torque acting on a dipole of dipole moment. (CBSE Delhi 2012)
Therefore the flux through the cube will be one-eighth of the total flux. Draw the pattern of electric field lines, when a point charge –Q is kept near an uncharged conducting plate. Two identical metallic spherical shells A and B having charges $+4\; Q$ and $ 10\; Q$ are kept a certain distance apart. = \(\vec{E} \cdot \overrightarrow{d s}_{1}+\vec{E} \cdot \overrightarrow{d s}_{2}+\vec{E} \cdot \overrightarrow{d s}_{3}\), = Eds1 cos 180° + Eds2 cos 90° + Eds3 cos 0° The magnitude of the force exerted on the second object by the first is F. If the first object is removed and replaced with an identical object that carries a charge +4q, what is the magnitude of the electric force on the second object?
What is the amount of work done in moving a point charge Q. around a circular arc of radius ‘r’ at the centre of which another point charge ‘q’ is located? Answer: [1996-3 marks] Ans. (CBSE Delhi 2018) E = \(\sqrt{3} E_{B}\) = \(\sqrt{3} \frac{k q}{a^{2}}\). Therefore, Answer: Question 21. Is it a scalar or vector? (Given K water = 80).
Draw electric field lines due to (i) two similar charges, (ii) two opposite charges, separated by a small distance. we have Work done in moving the charge “q”. Which spheres are they, if the final charge on each one is + 5.0 µC?
From point P, draw a Gaussian surface which will be a spherical shell of radius r. Let dS be a small area element on the Gaussian surface P. The radii of two metallic spheres A and B are r 1 and r 2 respectively (r 1 > r 2).
= 50 × 2 × 25 × 10-4 = 0.250 N m2 C-1, Therefore net flux is – 0.125 + 0.250 = + 0.125 N m2 C-1, (ii) Charge enclosed by the cylinder.
There are three unchanged identical metallic spheres `1,2 and 3` each of radius `r` and are placed at the verticles of an equlilateral triangle of side `d`.A charged metallic sphere having equal `q` of same radius `r` is touched to sphere `1` after some time it is taken to the location of sphere `2` and is touched to it ,then it is taken far away form sphere `1,2,3` after that the sphere `3 . i.
Two fixed point charges +4e and +e units are separated by a distance ‘a’. Charge on C after contact with A
Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +4q. (a) Use Gauss’s theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density σ. (CBSE Al 2017) Or It strikes the balance between the rigorousness of the Callen text and phenomenological approach of the Atkins text. The book is divided in three parts.
Answer: 13.
A metallic spherical shell has an inner radius R1 and outer radius R2. Example.
i. e. the electric field outside the spherical shell behaves as if the whole charge is concentrated at the center of the spherical shell. Answer: Let E and EB be the electric fields at point P due to the charges at A and B respectively.
No, the field may be normal to the surface. Calculate the linear charge density. Answer: (ii) Two identical metallic spheres A and B having charges + 4 Q and − 1 0 Q are kept a certain distance apart, a third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B, then, spheres A and B are brought in contact and then separated.
E = \(\frac{\sigma}{2 \varepsilon_{0}}\) , electric field due to an infinite plane sheet of charge.
τ = either force × arm of the couple Two identical charged spheres suspended from a common point by two massless string of lengths l, are initially at a distance d(d < < l) apart because of their mutual repulsion.
(a) The diagram is as shown.
Charge on earthed sphere is The third charge Q. wilt is in equilibrium if it experiences zero net force.
When the dipole is placed perpendicular to the field, two forces acting on the dipole form a couple, and hence a torque acts on it which aligns its dipole along the direction of the electric field. Calculate the electric dipole moment of the system. (b) Given the electric field in the region E = 2xî, find the net electric flux through the cube and the charge enclosed by it. This is the standalone version of University Physics with Modern Physics, Twelfth Edition.
19. The outer surface of the shell. Question 17. The drawing shows three point charges fixed in place. (CBSE Ai 2019)
E .
A solid metallic sphere has a charge + 3 Q. CiConcentric wihith this sphere is a conducting spherical shell having charge ‐Q. A third identical charged sphere C is first placed in contact with sphere A and then with sphere B, then spheres A and B are brought in contact and then separated. (a) its plane is parallel to the Y – Z plane, and Find the value of an electric field that would completely balance the weight of an electron. Electric field intensity at a point is defined as the force experienced by a unit test charge placed at that point. Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole. What would be the effect on the force when a plastic sheet is inserted between the two?
This field is in the direction of the dipole moment. ΦR = \(\vec{E}\).\(\vec{ds}\) = EdS cos 0°
Force of repulsion between sphere A having charge q/2 and sphere B having charge Therefore, the force of attraction between the two spheres is 5.703 × 10 - 3 N. Q14 : Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Electric flux: It is the measure of the number of electric field lines crossing a given area normally.
Justify each answer. 32 PHY2049Fall2015-$Acosta,Woodard$ $ Exam1$solutions$ Problem4!
What is the minimum value of the potential energy? Charge on outer surface + Q, Answer: Four charges of came magnitude and same sign are placed at the corners of a square, of each side 0.1 m. What is electric field intensity at the center of the square? Φ = \(\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}=\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\) … (2), from equations (1) and (2) it follows that
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