What will be the total flux through
Question: What will be the total flux through the faces of the cube (Fig. 1.9) with side of lengthaif a chargeqis placed at (a) A: a corner of the cube. (b) B: mid-point of an edge of the cube. (c) C: centre of a face of the cube. (d) D: mid-point of B and C. Solution: (a) When charge is placed at corner A of the cube then only 1/8thportion of the charge lies incised the Gaussian surface. So, total flux through the faces of the given cube =q/8o. (b) When charge is placed at point B of the cube t...
Read More →Sketch the electric field lines for a uniformly
Question: Sketch the electric field lines for a uniformly charged hollow cylinder shown in Fig 1.8. Solution:...
Read More →If the total charge enclosed by a surface is zero,
Question: If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero. Solution: According to the Gausss law, $\int_{S} E . d s=\frac{q}{\varepsilon_{o}}$ Here, the termqon the right side of the equation includes the sum of all charges enclosed by the surface (irrespective of the position of the charges inside the surface)...
Read More →The dimensions of an atom are of the order
Question: The dimensions of an atom are of the order of an Angstrom. Thus there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero? Solution: In an atom, number of electrons and protons are equal and the electric fields bind the atoms to neutral entity. Fields are caused by excess charges. There can be no excess charge on the inter surface of an isolated conductor....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \tan ^{5} x \sec ^{4} x d x$ Solution: Let $I=\int \tan ^{5} x \sec ^{4} x d x$ $\Rightarrow I=\int \tan ^{5} x \sec ^{2} x \sec ^{2} x d x$ $\Rightarrow I=\int \tan ^{5} x\left(1+\tan ^{2} x\right) \sec ^{2} x d x$ $\Rightarrow I=\int\left(\tan ^{5} x+\tan ^{7} x\right) \sec ^{2} x d x$ Let $\tan x=t$, then $\Rightarrow \sec ^{2} x d x=d t$ $\Rightarrow I=\int\left(t^{5}+t^{7}\right) d t$ $\Rightarrow I=\frac{t^{6}}{6}+\frac{t^{8}}{8}+c$ $\Right...
Read More →A metallic spherical shell has an inner radius R1
Question: A metallic spherical shell has an inner radius R1and outer radius R2. A charge Q is placed at the centre of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface? Solution: The situation given in question is shown in figure given below When positive charge is placed at the centre of the spherical cavity then an equal amount of negative charge (‒Q) appears on inner surface of the sphere due to induction. This charge is distributed...
Read More →An arbitrary surface encloses a dipole.
Question: An arbitrary surface encloses a dipole. What is the electric flux through this surface? Solution: From Gauss' law, the electric flux through an enclosed surface is given by $\int_{S} E \cdot d s=\frac{q}{\varepsilon_{o}}$ Where,qis the net charge enclose by the surface Now total charge enclose by the dipole = (‒q+q) = 0 Therefore, electric flux through a surface enclosing a dipole = 0....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \tan x \sec ^{4} x d x$ Solution: Let $I=\int \tan x \sec ^{4} x d x$ $\Rightarrow I=\int \tan x \sec ^{2} x \sec ^{2} x d x$ $\Rightarrow I=\int \tan x\left(1+\tan ^{2} x\right) \sec ^{2} x d x$ $\Rightarrow I=\int\left(\tan x+\tan ^{3} x\right) \sec ^{2} x d x$ Let $\tan x=t$, then $\Rightarrow \sec ^{2} x d x=d t$ $\Rightarrow I=\int\left(t+t^{3}\right) d t$ $\Rightarrow I=\frac{t^{2}}{2}+\frac{t^{4}}{4}+c$ $\Rightarrow I=\frac{\tan ^{2} x}{2}...
Read More →A positive charge Q is uniformly distributed along
Question: A positive chargeQis uniformly distributed along a circular ring of radiusR. A small test chargeqis placed at the centre of the ring (Fig. 1.7). Then (a) Ifq 0 and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre. (b) Ifq 0 and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring. (c) Ifq 0, it will perform SHM for small displacement along the ax...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \tan ^{3} x \sec ^{2} x d x$ Solution: Let $I=\int \tan ^{3} x \sec ^{2} x d x$ Let $\tan x=t$, then $\Rightarrow \sec ^{2} x d x=d t$ $\Rightarrow I=\int t^{3} d t$ $\Rightarrow I=\frac{t^{4}}{4}+c$ $\Rightarrow I=\frac{\tan ^{4} x}{4}+c$ Therefore, $\int \tan ^{3} x \sec ^{2} x d x=\frac{\tan ^{4} x}{4}+c$...
Read More →Refer to the arrangement of charges in
Question: Refer to the arrangement of charges in Fig. 1.6 and a Gaussian surface of radius R with Q at the centre. Then (a) total flux through the surface of the sphere is ‒Q/o (b) field on the surface of the sphere is ‒Q/(4oR2) (c) flux through the surface of sphere due to 5Qis zero (d) field on the surface of sphere due to 2Qis same everywhere Solution: Gauss' law states that total electric flux of an enclosed surface is given byq/o. Here,qis the net charge enclosed by the Gaussian surface. Fr...
Read More →Consider a region inside which there are various types
Question: Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region, (a) the electric field is necessarily zero (b) the electric field is due to the dipole moment of the charge distribution only (c) the dominant electric field is 1/r3 for larger, whereris the distance from a origin in this regions (d) the work done to move a charged particle along a closed path, away from the region, will be zero Solution: Here, net charge is zer...
Read More →If there were only one type
Question: If there were only one type of charge in the universe, then (a) $\oint_{z}$ E.ds $\neq 0$ on any surface (b) $\oint_{z} E \cdot d s=0$ if the charge is outside the surface (c) $\oint_{s} E . \mathrm{d} S$ could not be defined (d) $\oint_{s} \mathrm{E} \cdot \mathrm{dS}=\frac{q}{\varepsilon_{0}}$ if charges of magnitude $q$ were inside the surface Solution: According to the Gauss' law $\int_{S} E \cdot d s=\frac{q}{\varepsilon_{o}}$ Whereqis the charge enclosed by the surface. If thecha...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{x^{1 / 3}\left(x^{1 / 3}-1\right)} d x$ Solution: Let I $=\int \frac{1}{x^{\frac{1}{3}}\left(x^{\left.\frac{1}{3}-1\right)}\right.} d x$ Multiplying and dividing by $x^{\frac{1}{3}}$ $\Rightarrow I=\int \frac{x^{\frac{1}{3}}}{x^{\frac{2}{3}}\left(x^{\frac{1}{3}}-1\right)} d x$ Let, $x^{\frac{1}{3}}-1=t \Rightarrow \frac{1}{3} x^{-\frac{2}{3}} d x=d t$ So, $\Rightarrow I=3 \int \frac{(t+1)}{t} d t$ $\Rightarrow I=3 \int\left(t+\frac{1}{t}...
Read More →The electric field at a point is
Question: The electric field at a point is (a) always continuous (b) continuous if there is no charge at that point (c) discontinuous only if there is a negative charge at that point (d) discontinuous if there is a charge at that point Solution: As we know, electric field lines emanate from positive charges and terminate on negative charges. Thus, electric field due to the charge Q will be continuous, if there is no charge at that point. It will be discontinuous if there is a charge at that poin...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} \mathrm{dx}$ Solution: Let $I=\int \frac{1}{\sqrt{x}+\sqrt[4]{x}} d x$ $\Rightarrow I=\int \frac{1}{\sqrt[4]{x}(\sqrt[4]{x}+1)} d x$ Multiplying and dividing by $\sqrt{x}$ $\Rightarrow I=\int \frac{x^{\frac{1}{2}}}{x^{\frac{3}{4}}(\sqrt[4]{x}+1)} d x$ Let $\sqrt[4]{\mathrm{x}}+1=\mathrm{t} \Rightarrow \frac{1}{4} \mathrm{x}^{-\frac{3}{4}} \mathrm{dx}=\mathrm{dt}$ So, $\Rightarrow I=4 \int \frac{(t-1)^{2}}{t} d t$ $\...
Read More →Prove the following
Question: If$\int_{s} E \cdot d s=0$ over a surface, then (a) the electric field inside the surface and on it is zero (b) the electric field inside the surface is necessarily uniform (c) the number of flux lines entering the surface must be equal to the number of flux lines leaving it (d) all charges must necessarily be outside the surface Solution: Given, $\int_{s} E \cdot d s=0$ It means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int x(1-x)^{23} d x$ Solution: Let I $=\int \mathrm{x}(1-\mathrm{x})^{23} \mathrm{dx}$ Substituting $1-x=t \Rightarrow d x=-d t$ $\Rightarrow I=-\int(1-t) t^{23} d t$ $\Rightarrow I=-\int\left(t^{23}-t^{24}\right) d t$ $\Rightarrow I=-\left[\frac{t^{24}}{24}-\frac{t^{25}}{25}\right]+c$ $\Rightarrow I=\frac{t^{25}}{25}-\frac{t^{24}}{24}+c$ $\Rightarrow I=\frac{(1-x)^{25}}{25}-\frac{(1-x)^{24}}{24}+c$ $\Rightarrow I=\frac{1}{600}(1-x)^{24}[24(1-x)-25]$...
Read More →A point charge +q, is placed at a distance d
Question: A point charge +q, is placed at a distancedfrom an isolated conducting plane. The field at a pointPon the other side of the plane is (a)directed perpendicular to the plane and away from the plane. (b)directed perpendicular to the plane but towards the plane. (c)directed radially away from the point charge. (d)directed radially towards the point charge. Solution: When you place a positive charge near a conducting plane, then electric field lines from positive charges will enter into the...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int x(1-x)^{23} d x$ Solution: Let I $=\int \mathrm{x}(1-\mathrm{x})^{23} \mathrm{dx}$ Substituting $1-x=t \Rightarrow d x=-d t$ $\Rightarrow I=-\int(1-t) t^{23} d t$ $\Rightarrow I=-\int\left(t^{23}-t^{24}\right) d t$ $\Rightarrow I=-\left[\frac{t^{24}}{24}-\frac{t^{25}}{25}\right]+c$ $\Rightarrow I=\frac{(1-x)^{25}}{25}-\frac{(1-x)^{24}}{24}+c$ $\Rightarrow I=\frac{1}{600}(1-x)^{24}[24(1-x)-25]$ $\Rightarrow I=-\frac{1}{600}(1-x)^{24}[1+24 x]+c$...
Read More →Figure 1.5 shows electric field lines in
Question: Figure 1.5 shows electric field lines in which an electric dipolepis placed as shown. Which of the following statements is correct? (a)The dipole will not experience any force. (b)The dipole will experience a force towards right. (c)The dipole will experience a force towards left. (d)The dipole will experience a force upwards. Solution: We know electric field emerges radially outward from positive point charge. In the figure given above, space between field lines is increasing (or dens...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x^{2}}{\sqrt{1-x}} d x$ Solution: $\operatorname{Let} I=\int \frac{x^{2}}{\sqrt{1-x}} d x$ Substituting $1-x=t \Rightarrow d x=-d t$, $\Rightarrow \mathrm{I}=-\int \frac{(1-\mathrm{t})^{2}}{\sqrt{\mathrm{t}}} \mathrm{dt}$ $\Rightarrow \mathrm{I}=-\int \frac{\mathrm{t}^{2}-2 \mathrm{t}+1}{\sqrt{\mathrm{t}}} \mathrm{dt}$ $\Rightarrow \mathrm{I}=-\int\left(\mathrm{t}^{\frac{3}{2}}-2 \mathrm{t}^{\frac{1}{2}}+\mathrm{t}^{-\frac{1}{2}}\right) \ma...
Read More →Five charges q1, q2, q3, q4, and q5
Question: Five chargesq1,q2,q3,q4, andq5are fixed at their positions as shown in Fig. 1.4. S is a Gaussian surface. The Gausss law is given by Which of the following statements is correct? (a)Eon the LHS of the above equation will have a contribution fromq1,q5andq3whileqon the RHS will have a contribution fromq2andq4only. (b)Eon the LHS of the above equation will have a contribution from all charges whileqon the RHS will have a contribution fromq2andq4only. (c)Eon the LHS of the above equation w...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$ Solution: $\operatorname{Let} I=\int \frac{x^{2}+3 x+1}{(x+1)^{2}} d x$ Substituting $x+1=t \Rightarrow d x=d t$ $\Rightarrow I=\int \frac{(t-1)^{2}+3(t-1)+1}{t^{2}} d t$ $\Rightarrow I=\int \frac{t^{2}-2 t+1+3 t-3+1}{t^{2}} d t$ $\Rightarrow I=\int \frac{t^{2}+t-1}{t^{2}} d t$ $\Rightarrow I=\int\left(1+\frac{1}{t}-\frac{1}{t^{2}}\right) d t$ $\Rightarrow \mathrm{I}=\mathrm{t}+\log |\mathrm{t}|-\frac{1}{\mathrm...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int\left(2 \mathrm{x}^{2}+3\right) \sqrt{\mathrm{x}+2} \mathrm{dx}$ Solution: Let $\mathrm{I}=\int\left(2 \mathrm{x}^{2}+3\right) \sqrt{\mathrm{x}+2} \mathrm{~d} \mathrm{x}$ Substituting $x+2=t \Rightarrow d x=d t$ $\Rightarrow I=\int\left[2(t-2)^{2}+3\right] \sqrt{t} d t$ $\Rightarrow I=\int\left[2 t^{2}-8 t+8+3\right] \sqrt{t} d t$ $\Rightarrow I=\int\left[2 t^{\frac{5}{2}}-8 t^{\frac{3}{2}}+11 \frac{1}{2}\right] d t$ $\Rightarrow I=\frac{4}{7} t^{...
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