A sphere of radius R has a uniform volume charge density \rho . A spherical cavity of radius b whose centre lies at \vec {r} = \vec {a} is removed from the sphere.a. Find the electric field at any point inside the spherical cavity.b. Find the electric field outside the cavity.

a. The field within the cavity or outside is the superposition of the field due to the original uncut sphere, plus the field due to a sphere of the size of the cavity but with a uniform negative charge density. The effective charge distribution is composed of a uniformly charged sphere of radius r , charge density \rho , superposed on it, a charge density -\rho filling the cavity. Electric field \vec {E}_{1} caused by the charge distribution +\rho at a point \vec {r} ; inside the spherical cavity. \vec {E}_{1} = \dfrac {\rho r}{3\epsilon_{0}} \hat {r} = \dfrac {\rho \vec {r}}{3\epsilon_{0}} ; where \hat {r} is a unit vector in radial direction.Similarly, the electric field \vec {E}_{2} formed by the charge density -\rho inside the cavity is \vec {E}_{2} = \dfrac {(-\rho)\vec {s}}{3\epsilon_{0}}; \vec {s} = \vec {r} - \vec {a} \vec {s} is the radius vector from cavity centre to the point P . \therefore \vec {E}_{2} = \dfrac {-\rho (\vec {r} - \vec {a})}{3\epsilon_{0}} The resultant electric field inside the cavity is, therefore, given by the superposition of \vec {E}_{1} and \vec {E}_{2} , i.e. \vec {E} = \vec {E}_{1} + \vec {E}_{2} = \dfrac {\rho \vec {r}}{3\epsilon_{0}} + \left [\dfrac {-\rho (\vec {R} - \vec {a})}{3\epsilon_{0}}\right ] = + \dfrac {\rho \vec {a}}{3\epsilon_{0}} = constant \Rightarrow \vec {E} = \dfrac {\rho \vec {a}}{3\epsilon_{0}} .b. (i) Electric field at points inside the large sphere but outside the cavity, \vec {E} = \dfrac {\rho \vec {r}}{3\epsilon_{0}} and \vec {E}_{2} = \dfrac {1}{4\pi \epsilon_{0}} \dfrac {q\vec {s}}{s^{3}} = \dfrac {\left (-\dfrac {4}{2} \pi \rho b^{3}\right )(\vec {R} - \vec {a})}{4\pi \epsilon_{0} |\vec {r} - \vec {a}|^{3}} The resultant electric field is, therefore, \vec {E} = \vec {E}_{1} + \vec {E}_{2} = \dfrac {\rho}{3\epsilon_{0}} \left [\vec {r} - \left (\dfrac {b}{|\vec {r} - \vec {a}|}\right )(\vec {r} - \vec {a})\right ] (ii) Electric field at points outside the large sphere, \vec {E}_{1} = \dfrac {Q_{total}}{4\pi \epsilon_{0}r^{3}} \vec {r} = \dfrac {\left (\dfrac {4}{3} \pi R^{3}\rho \right )}{4\pi \epsilon_{0} r^{3}} \vec {r} = \dfrac {R^{3}\rho}{3\epsilon_{0}r^{3}} \vec {r} \vec {E}_{2} = \dfrac {q}{4\pi \epsilon_{0}r^{3}} \vec {r} = \dfrac {\left (-\dfrac {4}{3}\pi b^{3} \rho \right )}{4\pi \epsilon_{0}(|\vec {r} - \vec {a}|)^{3}} (\vec {r} - \vec {a}) = \dfrac {-\rho b^{3}}{3\epsilon_{0}(|\vec {r} - \vec {a}|)^{3}} (\vec {r} - \vec {a}) The resultant electric field, \vec {E} = \vec {E}_{1} + \vec {E}_{2} = \dfrac {\rho}{3\epsilon_{0}} \left [\left (\dfrac {R}{r}\right )^{3} \vec {r} - \left (\dfrac {b}{|\vec {r} - \vec {a}|}\right )^{3} (\vec {r} - \vec {a})\right ] E(\vec {r}) = \left\{\begin{matrix}\dfrac {\rho \vec {a}}{3\epsilon_{0}}\text {Electric field inside the cavity}\\ \dfrac {\rho}{3\epsilon_{0}} \left [\vec {r} = \left (\dfrac {b}{|\vec {r} - \vec {a}|}\right )^{3} (\vec {r} - \vec {a})\right ]\\ \text {Electric field outside the cavity but inside the large sphere}\\ \dfrac {\rho}{3\epsilon_{0}} \left [\left (\dfrac {R}{r}\right )^{3} \vec {r} - \left (\dfrac {b}{|\vec {r} - \vec {a}|}\right )^{3} (\vec {r} - \vec {a})\right ]\\ \text {Electric field outside the large sphere}\end{matrix}\right. .

Electric Charges And Fields - Difficult Conceptual Numericals

Q: Find the electric flux crossing the wire frame ABCD of length l, width b, and center at a distance OP=d from an infinite line of charge with linear charge density \lambda . Consider that the plane of the frame is perpendicular to the line OP Fig.

The flux passing through the strip of area dA is d \phi= EdA cos \theta =\dfrac{\lambda}{2 \pi \varepsilon_0\sqrt{d^2+x^2}}(l dx) \dfrac{d}{\sqrt{d^2+x^2}} d\phi=\dfrac{\lambda dl}{2 \pi \varepsilon_0} \dfrac{d}{(\sqrt{d^2+x^2})} \therefore \phi =\int d \phi=\dfrac{\lambda dl}{2 \pi \varepsilon_0} \int_{-b/2}^{b/2}\dfrac{d}{d^2+x^2} =\dfrac{\lambda dl}{2 \pi \varepsilon_0} \dfrac{1}{d}\left [ tan^{-1} \dfrac{x}{d} \right ]^{b/2}_{-b/2} =\dfrac{N}{2\pi \varepsilon_0}\times 2tan^{-1}\left ( \dfrac{b}{2d} \right ) \phi=\dfrac{N}{2\pi \varepsilon_0} tan^{-1}\left ( \dfrac{b}{2d} \right )

Q: Four point charges of charge Q are placed on the vertices of square and body of mass m and charge q is placed perpendicular to the centre of sqaure at a distance h from the centre. Take the distance between centre and vertices of the sqaure to be `a' . What should be the value of Q in order that this body may be in equilibrium?

O is the center of square and point P is at height h from the center OA = OB = OC = OD = a The magnitude of the electric force due to each charge is \displaystyle \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{a^2 + h^2} The components of the the force perpendicular to OP sum up to zero because of the symmetrical distribution of charges about OP. Hence, the resultant force at P is upward along OP. The magnitude of the force is given by \displaystyle R_{up} = 4 \times \frac{1}{4 \pi \varepsilon_0} \frac{Qq}{h^2 + a^2} cos \theta = \displaystyle \frac{Qq}{\pi \varepsilon_0 (h^2 + a^2)} \frac{h}{\sqrt{h^2 + a^2}} = \frac{Qqh}{\pi \varepsilon_0 (h^2 + a^2)^{3/2}} F_{down} = weight = mg For equilibrium, \displaystyle mg = \frac{Qqh}{\pi \varepsilon_0 (h^2 + a^2)^{3/2}} or \displaystyle Q = \pi \varepsilon_0 \frac{mg}{hq} (h^2 + a^2)^{3/2}

Q: A bob of a simple pendulum of mass 40 gm with a positive charge 4\times 10^{-6} is oscillating with a time period T_{1} .An electric field of intensity 3.6\times 10^{4}N/C is applied vertically upwards.Now the time period is T_{2} the value of \dfrac{T_{2}}{T_{1}} is \left ( g=10m/s^{2} \right ) :

We know that time period of pendulum is given by T_{1}=2 \pi \sqrt{\dfrac{l}{g}} when E will be applied, effective g will becomes g-\dfrac{Eq}{m} \Rightarrow T_{2}=2 \pi \sqrt{\left(\dfrac{l}{g-\dfrac{Eq}{m}}\right)} \dfrac{T_{2}}{T_{1}}=\sqrt{\dfrac{g}{g-\dfrac{Eq}{m}}} \dfrac{T_{2}}{T_{1}}=\sqrt{\dfrac{10}{10-\dfrac{3.6 \times 10^{4} \times 4 \times 10^{-6}}{40 \times 10^{-3}}}} \dfrac{T_{2}}{T_{1}}=\sqrt{\dfrac{10}{6.4}} \dfrac{T_{2}}{T_{1}}=\dfrac{10}{8} \dfrac{T_{2}}{T_{1}}=1.25

Q: Find the force experienced by a semicircular rod having a charge q as shown in figure. Radius of the wire is R, and the line of charge with linear charge density \lambda passes through its center and is perpendicular to the plane of wire.

\displaystyle F_{net} = \int dq E cos \theta = \displaystyle \int_{- \pi/2}^{\pi/2}\left ( \frac{q}{\pi R} \right ) Rd \theta \frac{\lambda}{2 \pi \varepsilon \cdot R} cos \theta = \displaystyle \frac{\lambda q}{2 \pi^2 \varepsilon_0 R} \int_{- \pi/2}^{\pi/2} cos \theta d \theta = \frac{\lambda q}{2 \pi^2 \varepsilon_0 R} [sin \theta]_{- \pi/2}^{\pi/2} = \displaystyle \frac{\lambda q}{2 \pi^2 \varepsilon_0 R} [1 - (-1)] = \frac{\lambda q}{\pi^2 \varepsilon_0 R}

Q: An infinite no.of electric charges each equal to 5\ nC are placed along X-axis at x = 1 \ cm, x = 2cm, x = 4\ cm, x=8\ cm and so on. In this setup, if the consecutive charges have opposite sign, then the electric field in newton/coulomb at x = 0 is: \left (\dfrac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 Nm^2/c^2 \right)

E due to all positive charges will be in -\hat x direction and due to all negative charges will be in +\hat x directionE due to positive =\dfrac {k(5\times 10^{-9})}{(1\times 10^{-2})^2}+\dfrac {k(5\times 10^{-9})}{(4\times 10^{-2})^2}+\dfrac {k(5\times 10^{-9})}{(16\times 10^{-2})^2}+---- =k(5\times 10^{-9})\dfrac {\dfrac {1}{(1\times 10^{-2})^2}}{1-\dfrac {1}{16}}=\dfrac {(9\times 10^9)(5\times 10^{-9})}{10^{-4}}\times \dfrac {16}{15}=48\times 10^4(-\hat x) E due to negative =\dfrac {k(5\times 10^{-9})}{(2\times 10^{-2})^2}+\dfrac {k(5\times 10^{-9})}{(8\times 10^{-2})^2}+---- =\dfrac {k(5\times 10^{-9})}{10^{-4}}(\dfrac {\dfrac {1}{4}}{1-\dfrac {1}{16}})=12\times 10^4(\hat x) \therefore E_{net}=(48-12)\times 10^4 \therefore E_{net}=36\times 10^4(-\hat x)\ N/C

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Difficult Conceptual Numericals

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Electric Charges And Fields

Find the electric flux crossing the wire frame ABCD of length l, width b, and center at a distance OP=d from an infinite line of charge with linear charge density

λ

. Consider that the plane of the frame is perpendicular to the line OP Fig.

\frac{2 N}{π ε _{0}} t a n^{- 1} (\frac{b}{2 d})

\frac{N}{π ε _{0}} t a n^{- 1} (\frac{b}{2 d})

\frac{3 N}{π ε _{0}} t a n^{- 1} (\frac{7 b}{2 d})

\frac{N}{2 π ε _{0}} t a n^{- 1} (\frac{b}{2 d})

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Four point charges of charge

Q

are placed on the vertices of square and body of mass m and charge q is placed perpendicular to the centre of sqaure at a distance

h

from the centre. Take the distance between centre and vertices of the sqaure to be

‘ a^{'}

. What should be the value of Q in order that this body may be in equilibrium?

π ε_{0} \frac{m g}{2 h q} (h^{2} + 2 a^{2})^{3 / 2}

π ε_{0} \frac{m g}{h q} (h^{2} + a^{2})^{3 / 2}

π ε_{0} \frac{2 m g}{h q} (h^{2} + 2 a^{2})^{3 / 2}

π ε_{0} \frac{m g}{2 h q} (h^{2} - a^{2})^{3 / 2}

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A bob of a simple pendulum of mass $40$ gm with a positive charge $4 \times 1 0^{- 6}$ is oscillating with a time period $T_{1}$ .An electric field of intensity $3.6 \times 1 0^{4} N / C$ is applied vertically upwards.Now the time period is $T_{2}$ the value of $\frac{T _{2}}{T _{1}}$ is $(g = 10 m / s^{2})$ :

0.16

0.64

1.25

0.8

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A sphere of radius

R

has a uniform volume charge density

ρ

. A spherical cavity of radius

b

whose centre lies at

r = a

is removed from the sphere.
a. Find the electric field at any point inside the spherical cavity.
b. Find the electric field outside the cavity.

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Find the force experienced by a semicircular rod having a charge q as shown in figure. Radius of the wire is R, and the line of charge with linear charge density

λ

passes through its center and is perpendicular to the plane of wire.

\frac{λ q}{2 π ^{2} ε _{0} R}

\frac{λ q}{π ^{2} ε _{0} R}

\frac{λ q}{4 π ^{2} ε _{0} R}

\frac{λ q}{8 π ^{2} ε _{0} R}

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An infinite no.of electric charges each equal to

5 n C

are placed along X-axis at

x = 1 c m, x = 2 c m, x = 4 c m, x = 8 c m

and so on. In this setup, if the consecutive charges have opposite sign, then the electric field in newton/coulomb at

x = 0

is:

(\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N m^{2} / c^{2})

12 \times 1 0^{4}

24 \times 1 0^{4}

36 \times 1 0^{4}

48 \times 1 0^{4}

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