Electric Charges And Fields

Q: Two charged balls of the same radius and weight suspended on threads of equal length are immersed into a liquid having density of d_{1} and a dielectric constant K . The density d of the material of the balls for the angles of divergence of the threads in the air and in the dielectric to be the same is:

When in air , for each ball , T cos\theta = Density \times volume \times g = dVg ( Vertical component of tension is balanced by weight ) T sin\theta = \dfrac{q^{2} }{4\pi \epsilon_0 ( 2l sin \theta) ^{2} } ( The horizontal component of tension is balanced by the Electrostatic Force ) tan\ \theta=\dfrac{\dfrac{q^{2}}{4\pi \epsilon_0 (2l\ sin\theta)^2}}{dVg} ----------- (1)Similarly ,when immersed in liquid , tan \theta = \dfrac{ \dfrac{ q^{2} }{4\pi K \epsilon_0 (2l sin \theta) ^{2} }}{Vg ( d-d_{1} )} ----------(2) ( Buoyant force, gravity and electrostatic force ) From (1) and (2) K(d-d_{1}) = d Hence , d = \dfrac{K d_{1} }{K-1}

Q: A charge +q is placed at (a, 0, 0) and another +q charge is placed at (-a, 0, 0) . A charge -q _1 is placed at the origin. If it is slightly displaced along y axis. Then :

Let the charge be displaced to (0,y) .The electric force due to a point charge is given by F = \displaystyle \frac{qQ}{4\pi\epsilon_0r^3}\vec{r} The electric force due to the charge at (a,0,0) is F_1 = \displaystyle \frac{1}{4\pi\epsilon_0}\frac{-qq_1}{(\sqrt{a^2+y^2})^3}(-a\hat{i}+y\hat{j}) The electric force due to the charge at (-a,0,0) is F_2 = \displaystyle \frac{1}{4\pi\epsilon_0} \frac{-qq_1}{(\sqrt{a^2+y^2})^3}(a\hat{i}+y\hat{j}) Thus, the total force experienced by q_1 due to the system of charges is F = F_1+F_2 = \displaystyle \frac{1}{4\pi\epsilon_0} \frac{-qq_1}{(\sqrt{a^2+y^2})^3}(2y\hat{j}) As the displacement is small, i.e., y<<a , F \approx \displaystyle \frac{1}{4\pi\epsilon_0} \frac{-qq_1}{a^3}(2 y)\hat{j} = -\frac{qq_1}{2\pi\epsilon_0a^3}y\hat{j} We get F \propto -y The force is proportional to the displacement and is in the opposite direction. Thus, it executes a SHM.

Q: A small electric dipole having dipole moment \vec {p} is placed along X-axis, as shown in the figure. A semi-infinite uniformly charged di-electric thin rod placed along x axis, with one end coinciding with origin. If linear charge density of rod is +\lambda and distance of dipole from rod is 'a', then calculate the electric force acting on dipole.

Electric field at A due to dipole, E_A = \dfrac{2KP}{r^3} where K = \dfrac{1}{4\pi \epsilon_o} E_A = \dfrac{2KP}{(x+a)^3} Force at A due to dipole, F_A = qE_A = (\lambda dx) \times \dfrac{2KP}{(x+a)^3} F_A = \dfrac{2\lambda KP}{(x+a)^3} dx Total force exerted due to dipole on the rod, F_D = \int^{\infty}_{0} \dfrac{2\lambda KP}{(x+a)^3} dx F_D = 2\lambda KP\times \dfrac{1}{-2(x+a)^2}\bigg|_{0}^{\infty} \implies F_A = \dfrac{\lambda KP}{a^2} Now force exerted by rod on dipole is equal and opposite to F_D (Newton's law) \vec{F_R} = - F_D = \dfrac{-\lambda P}{4\pi \epsilon_o a^2}

Q: A particle of mass 1 Kg carrying 0.01C charge is at rest on an inclined plane of angle 30^{0} with the horizontal when an electric field of \frac{490}{\sqrt{3}}NC^{-1} is applied parallel to the horizontal as shown. The coefficient of friction is :

For the body to be at rest on the incline, all the forces acting on it should balance out each other. Comparing forces along the incline , qE cos 30= \frac{490}{\sqrt{3}} \times \frac{\sqrt{3}}{2} \times 0.01 = 2.45 N mg sin 30 = 1 \times 9.8 \times \frac{1}{2} = 4.9 N \therefore qE cos 30 < mg sin 30 Hence , motion is downwards the plane and f is up the inclined plane .Force balance across the vertical direction gives mg = f sin \theta + N cos \theta where \theta = {30}^{o} Similarly, balancing forces along the x direction, N sin \theta = f cos \theta + qE Now, if \mu is the coefficient of friction, then f = \mu N Substituting the above values in the equations and solving them simultaneously, we get the value of \mu as \frac{\sqrt{3}}{7}

Q: An electron is projected, as shown in the figure, at a speed of 6\times 10^{6}ms^{-1} , at an angle of 45^{0} . It is given that E=2000Vm^{-1} is directed upward, d = 3cm and l=10 cm . The electron will strike at :

The acceleration due to the electric field E is \dfrac{eE}{m}= \dfrac{ 1.6 \times 10^{-19} \times 2000}{9.1 \times 10^{-31}}=\dfrac{3.2 \times 10^{15}}{9.1} downward and is constant.Hence, the motion is equivalent to a projectile motion. T = \dfrac{2u \sin \theta}{a} range R = \dfrac{u^2 \sin 2\theta}{a}=\dfrac{ ( 6 \times 10^6)^2 \sin( 2 \times 45^0)}{\dfrac{3.2 \times 10^{15}}{9.1}}=0.1m= 10 cm \therefore the electron will strike the lower plate at the edge.Max height reached: H = \dfrac{ u^2 \sin ^2 \theta}{2a} = \dfrac{ ( 6 \times 10^6)^2 \times ( \dfrac{1}{\sqrt{2}})^2}{2 \times \dfrac{3.2 \times 10^{15}}{9.1}}=0.025m=2.5\: cm Since max height reached by electron is less than the separation between the plates, it will not strike the upper plate. Hence it will touch the lower plate at the edge.

Q: A particle that carries a charge -q is placed at rest in uniform electric field 10\ N/C . It experiences a force and moves in a certain time t, it is observed to acquire a velocity 10\vec{i}-10\vec{j} m/s. The given electric field intersects a surface of area A m^{2} in the X -Z plane. Electric flux through surface is:

v=10\hat{i}-10\hat{j} As velocity is in x and y components Electric field also has two components Let E=Ex\hat{i}+Ey\hat{j} Now force on (-q)=E(-q) Force F=-Exq\hat{i}+-Eyq\hat{j} Acceleration a =(-\dfrac{Exq}{m})\hat{i} +(-\dfrac{EyV}{m})\hat{i} Now v_x=10 and v_y=-10 we know v =u+at \ \ (given \ u=0 ) \therefore V=at \Rightarrow v_x=(-\dfrac{Exq}{m})t --(1) \ \ and\ \ v_y=(\dfrac{-Eyq}{m})t --(2) Given |E|=10\ N/C Now from 1 and 2 we get, -1=\dfrac{-Exq}{m}\dfrac{m}{Eyq} \Rightarrow Ex=Ey 10=\sqrt{Ex^{2}+Ey^{2}} \Rightarrow Ex=5\sqrt{2}=Ey Now than \sigma = \int E.dA As the area is in X-Z plane Area = A \hat{i} \therefore Ea=Ey.A \phi=5\sqrt{2}A

Q: The electric field in a region is E = \displaystyle \frac{5\times 10^3 x}{2}\hat{i} \ NC^{-1} cm^{-1} . The charge contained inside a cubical volume bounded by the surfaces x = 0, x = 1, y = 0, y = 1, z = 0, z = 1 is (where x, y, z are in cm) :

Area of each surfaces a=1^2 cm^2=10^{-4} m^2 .As the field is in the direction of x-axis so only surfaces x=0 and x=1 will contribute the flux. For surface \displaystyle x=0, \phi_0=E.a=\frac{5\times 10^3\times x}{2}.a= \frac{5\times 10^3\times 0}{2}.a=0 and for \displaystyle x=1, \phi_1=E.a=\frac{5\times 10^3\times x}{2}.a=\frac{5\times 10^3\times 1}{2} \times 10^{-4}=0.25 By Gauss's law, \displaystyle \phi_1=\frac{q_{en}}{\epsilon_0} \Rightarrow q_{en}=0.25\times 8.85\times 10^{-12}=2.21 \times 10^{-12} C

Q: A system consists of a ball of radius R carrying a uniformly distributed charge q and surrounding space filled with a charge of volume density \rho = \alpha / r where \alpha is a constant and r is the distance from the centre of the ball. Find the charge on the ball for which the magnitude of electric field strength outside the ball is constant ? The dielectric constant of the ball and the surrounding may be taken equal to unity.

Consider a spherical surface of radius r(r< R) having center at the of the ball. If E is the magnitude of electric field strength at the surface, then the electric flux through this surface is \int_{s}\vec{E}\cdot d\vec{S}=\int_{s}EdS cos 0^{\circ}=E\int_{s}dS=E4 \pi r^2 By Gauss theorem, \int_{s}\vec{E}\cdot d\vec{S}=\frac{1}{\varepsilon_0}(Q_{enclosed}) If q is charge on ball, then Q_{enclosed}=q+ \int_{R}^{r} \rho dv=q+\int_{R}^{r}\frac{\alpha}{x}\cdot 4 \pi x^2 dx = q+ 2 \pi \alpha (r^2+R^2) E4 \pi r^2=\frac{1}{\varepsilon_0}[q+2 \pi \alpha (r^2-R^2)] =\dfrac{\alpha}{2\varepsilon_0}+\dfrac{1}{\varepsilon_0}\left ( \dfrac{q}{4 \pi r^2}-\dfrac{\alpha}{2}\dfrac{R^2}{r^2} \right ) This will be independent of r if the second term on RHS is zero, i.e., \dfrac{q}{4 \pi r^2}-\dfrac{\alpha R^2}{2r^2}=0 Then the electric field strength E will be \dfrac {\alpha}{2 \varepsilon_0} .

Q: A point charge placed on the axis of a uniformly charged disc experiences a force F due to the disc. If the surface charge density on the disc is \sigma , the electric flux through the disc due to the point charge will be :

The electric field due to the charged disc for an axial point, E_d=\dfrac{\sigma}{2\epsilon_0}(1-\dfrac{x}{\sqrt{x^2+R^2}})=\dfrac{\sigma}{2\epsilon_0}(1-\cos\theta_0) Force on q, F=qE_d=\dfrac{q\sigma}{2\epsilon_0}(1-\cos\theta_0) .....(1) Consider a ring of radius y and thickness dy . The flux through this ring, d\phi=(E\cos\theta)2\pi ydy (as only horizontal component contribute flux and flux due to vertical component is zero because E_y\hat{j}.\hat{i} =0 )Total flux, \phi=\int^{R}_0 (\dfrac{q}{4\pi\epsilon_0r^2}\cos\theta) 2\pi ydy \Rightarrow \phi=\dfrac{q}{2\epsilon_0}\int^{R}_0\dfrac{1}{x^2+y^2}\dfrac{x}{\sqrt{x^2+y^2}}ydy \phi=\dfrac{qx}{2\epsilon_0}\int^{R}_0\dfrac{y}{(x^2+y^2)^{3/2}}dy=\dfrac{q}{2\epsilon_0}[1-\dfrac{x}{\sqrt{x^2+R^2}}] \Rightarrow \phi=\dfrac{q}{2\epsilon_0}(1-\cos\theta_0) ....(2) From (1), (2) becomes, \phi=\dfrac{F}{\sigma}

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Difficult Questions

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- Are you well prepared? Practice hard questions to be more confident about the chapter

When two charged objects immersed in a dielectric liquid then the number of forces along with electrostatic force will come in picture. Let's practice this question.

Two charged balls of the same radius and weight suspended on threads of equal length are immersed into a liquid having density of $d_{1}$ and a dielectric constant $K$ . The density $d$ of the material of the balls for the angles of divergence of the threads in the air and in the dielectric to be the same is:

\frac{K d}{K - 1}

\frac{K - 1}{K d}

\frac{d}{K - 1}

\frac{K - 1}{d}

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A charge can do SHM in between two opposite charges after small displacement. Figure out the situation when this charge will do SHM.

A charge +q is placed at

(a, 0, 0)

and another

+ q

charge is placed at

(- a, 0, 0)

. A charge

- q

_{1}

is placed at the origin. If it is slightly displaced along y axis. Then :

it will move away

it will oscillate but will not execute SHM

it will execute SHM

it will stand at the displaced position

View solution

Electrostatic interaction between dipole and charged rod are quite complex. Force on the rod due to dipole can be calculated by integrating the value of force on a small length of rod 'dl' having charge $λ d l$ .

A small electric dipole having dipole moment

p

is placed along X-axis, as shown in the figure. A semi-infinite uniformly charged di-electric thin rod placed along x axis, with one end coinciding with origin. If linear charge density of rod is

+ λ

and distance of dipole from rod is 'a', then calculate the electric force acting on dipole.

\frac{- P λ}{2 π ϵ _{0} a ^{2}}

\frac{- 3 P λ}{4 π ϵ _{0} a ^{2}}

\frac{- 3 P λ}{2 π ϵ _{0} a ^{2}}

\frac{- P λ}{4 π ϵ _{0} a ^{2}}

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A charged block is maintaining equilibrium on a smooth inclined surface in presence of the horizontal electric field. In this situation, three different forces will act on the block at different-different angles. Let's practice this question and find the equilibrium condition and the direction of friction.

A particle of mass 1 Kg carrying 0.01C charge is at rest on an inclined plane of angle

3 0^{0}

with the horizontal when an electric field of

\frac{4 9 0}{3} N C^{- 1}

is applied parallel to the horizontal as shown. The coefficient of friction is :

0.5

\frac{1}{3}

\frac{3}{2}

\frac{3}{7}

View solution

Let's practice, the concepts of projectile motion of an electron in presence of an electric field.

An electron is projected, as shown in the figure, at a speed of

6 \times 1 0^{6} m s^{- 1}

, at an angle of

4 5^{0}

. It is given that

E = 2000 V m^{- 1}

is directed upward,

d = 3 c m

and

l = 10 c m

. The electron will strike at :

Upper plate

Mid point of the lower plate

Opposite end of the lower plate

No where

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A charged particle acquires some velocity from rest in presence of the electric field. Otherhand, In direction of the electric field, One surface is placed. Let's think how much flux will link with the given surface and solve the below question.

A particle that carries a charge $- q$ is placed at rest in uniform electric field $10 N / C$ . It experiences a force and moves in a certain time t, it is observed to acquire a velocity $10 i - 10 j$ m/s. The given electric field intersects a surface of area $A$ $m^{2}$ in the X -Z plane. Electric flux through surface is:

52 A N m^{2} / C

5 A N m^{2} / C

2 A N m^{2} / C

25 A N m^{2} / C

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Calculate the charge enclosed by the closed surface when it is kept in a linearly varying electric field.

The electric field in a region is

E = \frac{5 \times 1 0 ^{3} x}{2} \hat{i} N C^{- 1} c m^{- 1}

. The charge contained inside a cubical volume bounded by the surfaces

x = 0, x = 1, y = 0, y = 1, z = 0, z = 1

is (where x, y, z are in cm) :

2.21 \times 1 0^{- 12} C

4.42 \times 1 0^{- 12} C

2.21 \times 1 0^{- 8} C

4.42 \times 1 0^{- 8} C

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Find the condition of zero electric fields outside the ball when a ball has a fixed charge on its surface but in interior volume, charge distribution is varying with radial distance.

A system consists of a ball of radius

R

carrying a uniformly distributed charge

q

and surrounding space filled with a charge of volume density

ρ = α / r

where

α

is a constant and

r

is the distance from the centre of the ball. Find the charge on the ball for which the magnitude of electric field strength outside the ball is constant ? The dielectric constant of the ball and the surrounding may be taken equal to unity.

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A charged particle is applying force 'F' on the uniformly charged disc, then find the amount of electric flux linkage with the disc.

A point charge placed on the axis of a uniformly charged disc experiences a force F due to the disc. If the surface charge density on the disc is

σ

, the electric flux through the disc due to the point charge will be :

\frac{2 π F}{σ}

\frac{F}{2 π σ}

\frac{2 F}{σ}

\frac{F}{σ}

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