Q: A few electric field lines for a system of two charges \$Q_{1}\$ and \$Q_{2}\$ fixed at two different points on the \$x\$-axis are shown in the figure. These lines suggest that

Number of Electric field lines emerging from \$Q_1\$ are more than the number of lines terminating at \$Q_2\$.So , \$Q_1\$ is a positive charge of higher magnitude and \$Q_2\$ is a negative charge of smaller magnitude .Hence , the point at which the electric field is zero is on the right of the negative charge \$Q_2\$

Q: Consider a uniform electric field \$E=3\times 10^3 \hat i \: N/C\$. What is the flux of this field through a square of \$10\ cm\$ on a side whose plane is parallel to the yz plane?

The flux of an electric field is given by,\$ \phi = EA\$\$ \phi = 3 \times {10}^{3} \times 0.1 \times 0.1 \$\$ \phi = 30\ Nm^2/C \$

Q: In the figure shown, A is a fixed charged. B (of mass m) is given a velocity \$v\$ as shown in figure. At this moment the radius of curvature of the resultant path of B is :

\$F=\dfrac{kq^2}{r^2}=\dfrac{mv^2}{R_c}\$\$\Rightarrow R_c=\dfrac{mv^2}{F}=\dfrac{mv^2}{\dfrac{kq^2}{r^2}}=\dfrac{4\pi \varepsilon_0 r^2 mv^2}{q^2}\$

Q: A particle of charge \$-q\$ and mass \$m\$ moves in a circular orbit of radius \$r\$ about a fixed charge \$+Q\$. The relation between the radius of the orbit \$r\$ and the time period \$T\$ is:

The centrifugal force acting on charge \$q\$ is:\$F=\dfrac {k q Q}{r^2}\$\$\Rightarrow \dfrac {mv^2}{r}=\dfrac {k q Q}{r^2}\$\$\Rightarrow V=\sqrt {\dfrac {k q Q}{rm}}\$Now, \$V=\dfrac {2\pi r}{T}\$\$\Rightarrow \dfrac {2\pi r}{T}=\sqrt {\dfrac {q Q}{4\pi \varepsilon_0 r m}}\$\$\Rightarrow \dfrac {4\pi^2 r^2}{T^2}=\dfrac {q Q}{4\pi \varepsilon_0 r m}\$\$\Rightarrow r^3=\dfrac {q Q T^2}{16\pi^3 \varepsilon_0 m}\$

Q: Two point charges placed at a distance \$r\$ in air exert a force \$F\$ on each other. The value of distance \$R\$ at which they experience force \$4F\$ when placed in a medium of dielectric constant \$K = 9\$ is :

Initially force, \$F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2} .......(1)\$Finally, \$4F=\dfrac{1}{4\pi\epsilon}\dfrac{q_1q_2}{R^2}\$ where \$K=\dfrac{\epsilon}{\epsilon_0}\Rightarrow \epsilon=9\epsilon_0\$now,\$4F=\dfrac{1}{4\pi(9\epsilon_0)}\dfrac{q_1q_2}{R^2}......(2)\$\$(1)/(2) ,\dfrac{1}{4}=\dfrac{9R^2}{r^2}\Rightarrow R=\dfrac{r}{6}\$

Q: Electric flux through a surface of area \$100\ m^{2}\$ lying in the xy plane is (in V-m) if \$\vec{E}=\hat{i}+\sqrt{2}\hat{j}+\sqrt{3\hat{k}}\$

Given \$\overrightarrow E= \hat i+ \sqrt 2 \hat j+ \sqrt 3 \hat k\$, Area, \$\overrightarrow A= 100 \hat k\$Electric Flux\$= \overrightarrow E.\overrightarrow A\$\$\Rightarrow \text{Electric Flux}= (\hat i + \sqrt 2 \hat j + \sqrt 3 \hat k).( 100 \hat k)=100\sqrt 3=173.2\$

Q: Two similar very small conducting spheres having charges \$40 C\$ and \$ -20 C\$ are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

Initially force between conducting spheres \$F_1=\cfrac{Kq_1q_2}{r^2}=\cfrac{K(40)(-20)}{r^2}\$(i)After contact total charge\$=40-20=20\mu C\$Now charge will be equally distributedso \$q_1=q_2=\dfrac{20}{2}=10\mu C\$Finally force between spheres\$F_2 =\cfrac{K(10)(10)}{r^2} \$(ii)Ratio of forces=\$\cfrac{F_1}{F_2}=\cfrac{40\times 20}{10\times 10}=8\$

Q: A point charge \$q\$ is revolving in a circle of radius \$' r '\$ around a fixed infinite line charge with positive charge \$\lambda\$ per unit length. Now the point charge is shifted and it revolves in a circle of radius \$ 2 r \$. Then :

Electric field at distance \$r\$ from the infinite line charge is \$E=k \dfrac{2 \lambda}{r}\$The necessary centripetal force to the point charge of mass \$m\$ is provided by electrostatic attraction.\$\therefore \dfrac{mv^2}{r}=qE=qk \dfrac{2 \lambda}{r}\$ or \$v=\sqrt{\dfrac{2k\lambda q}{m}}\$The speed of charged particle is independent of orbital radius. Hence the K.E. of point charge in orbits of radius \$r\$ and \$2r\$ is same. As the work done is the change of K.E, so the work done by all forces is zero.

Question 1

A few electric field lines for a system of two charges $Q_{1}$ and $Q_{2}$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that

Accepted Answer

Number of Electric field lines emerging from $Q_1$ are more than the number of lines terminating at $Q_2$.So , $Q_1$ is a positive charge of higher magnitude and $Q_2$ is a negative charge of smaller magnitude .Hence , the point at which the electric field is zero is on the right of the negative charge  $Q_2$

Question 2

A hemispherical body is placed in a uniform electric field E. What is the flux linked with curved surface, if field is perpendicular to base in figure.

Accepted Answer

The flux entering the hemisphere is equal to the flux leaving it, Thus, $ \phi_{in} = \phi_{out} $$ \phi_{out} = \pi R^{2} E $

Question 3

Consider a uniform electric field $E=3\times 10^3 \hat i \: N/C$. What is the flux of this field through a square of $10\  cm$ on a side whose plane is parallel to the yz plane?

Accepted Answer

The flux of an electric field is given by,$ \phi = EA$$ \phi = 3 \times {10}^{3} \times 0.1 \times 0.1 $$ \phi = 30\ Nm^2/C $

Question 4

In the figure shown, A is a fixed charged. B (of mass m) is given a velocity $v$ as shown in figure. At this moment the radius of curvature of the resultant path of B is :

Accepted Answer

$F=\dfrac{kq^2}{r^2}=\dfrac{mv^2}{R_c}$$\Rightarrow R_c=\dfrac{mv^2}{F}=\dfrac{mv^2}{\dfrac{kq^2}{r^2}}=\dfrac{4\pi \varepsilon_0 r^2 mv^2}{q^2}$

Question 5

A particle of charge $-q$ and mass $m$ moves in a circular orbit of radius $r$ about a fixed charge $+Q$. The relation between the radius of the orbit $r$ and the time period $T$ is:

Accepted Answer

The centrifugal force acting on charge $q$ is:$F=\dfrac {k q Q}{r^2}$$\Rightarrow \dfrac {mv^2}{r}=\dfrac {k q Q}{r^2}$$\Rightarrow V=\sqrt {\dfrac {k q Q}{rm}}$Now, $V=\dfrac {2\pi r}{T}$$\Rightarrow \dfrac {2\pi r}{T}=\sqrt {\dfrac {q Q}{4\pi \varepsilon_0 r m}}$$\Rightarrow \dfrac {4\pi^2 r^2}{T^2}=\dfrac {q Q}{4\pi \varepsilon_0 r m}$$\Rightarrow r^3=\dfrac {q Q T^2}{16\pi^3 \varepsilon_0 m}$

Question 6

Two point charges placed at a distance $r$ in air exert a force $F$ on each other. The value of distance $R$ at which they experience force $4 F$ when placed in a medium of dielectric constant $K = 9$ is :

Accepted Answer

Initially force, $F=\dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2} .......(1)$Finally, $4F=\dfrac{1}{4\pi\epsilon}\dfrac{q_1q_2}{R^2}$  where $K=\dfrac{\epsilon}{\epsilon_0}\Rightarrow \epsilon=9\epsilon_0$now,$4F=\dfrac{1}{4\pi(9\epsilon_0)}\dfrac{q_1q_2}{R^2}......(2)$$(1)/(2)  ,\dfrac{1}{4}=\dfrac{9R^2}{r^2}\Rightarrow R=\dfrac{r}{6}$

Question 7

A positive point charge $+Q$ is fixed in space. A negative point charge $q$ of mass $m$ revolves around the fixed charge in an elliptical orbit. The fixed charge $+Q$ is at one focus of the ellipse. The only force acting on the negative charge is the electrostatic force due to the positive charge. Then which of the following statements is true?

Accepted Answer

Since net force on the negative charge is always directed towards fixed positive charge, the torque on negative charge about positive charge is zero. Therefore angular momentum of negative charge about the fixed positive charge is conserved.$L = (mr^2) \omega $$\therefore$ As value of r changes, $ \omega $ also changes

Question 8

Electric flux through a surface of area $100\ m^{2}$ lying in the xy plane is (in V-m) if $\vec{E}=\hat{i}+\sqrt{2}\hat{j}+\sqrt{3\hat{k}}$

Accepted Answer

Given $\overrightarrow E= \hat i+ \sqrt 2 \hat j+ \sqrt 3 \hat k$, Area, $\overrightarrow A= 100 \hat k$Electric Flux$= \overrightarrow E.\overrightarrow A$$\Rightarrow \text{Electric Flux}= (\hat i + \sqrt 2 \hat j + \sqrt 3 \hat k).( 100 \hat k)=100\sqrt 3=173.2$

Question 9

Two similar very small conducting spheres having charges $40 C$ and $- 20 C$ are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

Accepted Answer

Initially force between conducting spheres $F_1=\cfrac{Kq_1q_2}{r^2}=\cfrac{K(40)(-20)}{r^2}$(i)After contact total charge$=40-20=20\mu C$Now charge will be equally distributedso $q_1=q_2=\dfrac{20}{2}=10\mu C$Finally force between spheres$F_2 =\cfrac{K(10)(10)}{r^2} $(ii)Ratio of forces=$\cfrac{F_1}{F_2}=\cfrac{40\times 20}{10\times 10}=8$

Question 10

A point charge $q$ is revolving in a circle of radius $^{'} r^{'}$ around a fixed infinite line charge with positive charge $λ$ per unit length. Now the point charge is shifted and it revolves in a circle of radius $2 r$ . Then :

Accepted Answer

Electric field at distance $r$ from the infinite line charge is $E=k \dfrac{2 \lambda}{r}$The necessary centripetal force to the point charge of mass $m$ is provided by electrostatic attraction.$\therefore \dfrac{mv^2}{r}=qE=qk \dfrac{2 \lambda}{r}$ or $v=\sqrt{\dfrac{2k\lambda q}{m}}$The speed of charged particle is independent of orbital radius. Hence the K.E. of point charge in orbits of radius $r$ and $2r$ is same. As the work done is the change of K.E, so  the work done by all forces is zero.

Electric Charges And Fields

Physics
11145 Views

A few electric field lines for a system of two charges $Q_{1}$ and $Q_{2}$ fixed at two different points on the $x$ -axis are shown in the figure. These lines suggest that

A hemispherical body is placed in a uniform electric field E. What is the flux linked with curved surface, if field is perpendicular to base in figure.

Consider a uniform electric field $E = 3 \times 1 0^{3} \hat{i} N / C$ . What is the flux of this field through a square of $10 c m$ on a side whose plane is parallel to the yz plane?

In the figure shown, A is a fixed charged. B (of mass m) is given a velocity $v$ as shown in figure. At this moment the radius of curvature of the resultant path of B is :

A particle of charge $- q$ and mass $m$ moves in a circular orbit of radius $r$ about a fixed charge $+ Q$ . The relation between the radius of the orbit $r$ and the time period $T$ is:

Two point charges placed at a distance $r$ in air exert a force $F$ on each other. The value of distance $R$ at which they experience force $4 F$ when placed in a medium of dielectric constant $K = 9$ is :

Electric flux through a surface of area $100 m^{2}$ lying in the xy plane is (in V-m) if $E = \hat{i} + 2 \hat{j} + 3 \hat{k}$

Two similar very small conducting spheres having charges $40 C$ and $- 20 C$ are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

A point charge $q$ is revolving in a circle of radius $^{'} r^{'}$ around a fixed infinite line charge with positive charge $λ$ per unit length. Now the point charge is shifted and it revolves in a circle of radius $2 r$ . Then :

Electric Charges And Fields

Physics 11145 Views

A few electric field lines for a system of two charges Q1​ and Q2​ fixed at two different points on the x-axis are shown in the figure. These lines suggest that

A hemispherical body is placed in a uniform electric field E. What is the flux linked with curved surface, if field is perpendicular to base in figure.

Consider a uniform electric field E=3×103i^N/C. What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?

In the figure shown, A is a fixed charged. B (of mass m) is given a velocity v as shown in figure. At this moment the radius of curvature of the resultant path of B is :

A particle of charge −q and mass m moves in a circular orbit of radius r about a fixed charge +Q. The relation between the radius of the orbit r and the time period T is:

Two point charges placed at a distance r in air exert a force F on each other. The value of distance R at which they experience force 4F when placed in a medium of dielectric constant K=9 is :

Electric flux through a surface of area 100 m2 lying in the xy plane is (in V-m) if E=i^+2​j^​+3k^​

Two similar very small conducting spheres having charges 40C and −20C are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

A point charge q is revolving in a circle of radius ′r′ around a fixed infinite line charge with positive charge λ per unit length. Now the point charge is shifted and it revolves in a circle of radius 2r. Then :

Physics
11145 Views

A few electric field lines for a system of two charges $Q_{1}$ and $Q_{2}$ fixed at two different points on the $x$ -axis are shown in the figure. These lines suggest that

Consider a uniform electric field $E = 3 \times 1 0^{3} \hat{i} N / C$ . What is the flux of this field through a square of $10 c m$ on a side whose plane is parallel to the yz plane?

In the figure shown, A is a fixed charged. B (of mass m) is given a velocity $v$ as shown in figure. At this moment the radius of curvature of the resultant path of B is :

A particle of charge $- q$ and mass $m$ moves in a circular orbit of radius $r$ about a fixed charge $+ Q$ . The relation between the radius of the orbit $r$ and the time period $T$ is:

Two point charges placed at a distance $r$ in air exert a force $F$ on each other. The value of distance $R$ at which they experience force $4 F$ when placed in a medium of dielectric constant $K = 9$ is :

Electric flux through a surface of area $100 m^{2}$ lying in the xy plane is (in V-m) if $E = \hat{i} + 2 \hat{j} + 3 \hat{k}$

Two similar very small conducting spheres having charges $40 C$ and $- 20 C$ are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

A point charge $q$ is revolving in a circle of radius $^{'} r^{'}$ around a fixed infinite line charge with positive charge $λ$ per unit length. Now the point charge is shifted and it revolves in a circle of radius $2 r$ . Then :