Two charges q and - 3 q are placed fixed on x -axis separated by distance ^ { \prime }d ^ { \prime } . Where should a third charge 2q be placed such that it will not experience any force?

\textbf{Step 1: Representing charges and force on 2q} Let the charge 2q be placed at point P as shown (at a distance x from q ) \textbf{Step 2: Forces on charge 2q:} F_{-3q}: Force due to charge -3q on 2q F_{-3q} = \dfrac{K 3q \times 2q}{(x + d)^2} = \dfrac{K 6q^2}{(x + d)^2} F_q= Force due to charge q on 2q F_q = K \dfrac{2q \times q}{x^2} = K \dfrac{2q^2}{x^2} \textbf{Step 3: Net force on charge 2q:} Net force on charge 2q is zero. \therefore F_{-3q} = F_q \dfrac{K6q^2}{(x + d)^2} = K \dfrac{2q^2}{x^2} 3x^2 = (x + d)^2 3x^2 = x^2 + d^2 + 2 xd 2x^2 - 2xd - d^2 = 0 The roots of this quadratic equation are x = \dfrac{2d \pm \sqrt{4d^2 + 8d^2}}{4} = \dfrac{d \pm \sqrt{3} d}{2} x = \dfrac{d - \sqrt{3}d}{2} is not acceptable. \therefore x = \dfrac{d + \sqrt{3} d}{2} = d \left(\dfrac{1 + \sqrt{3}}{2} \right) \therefore Charge 2q must be placed at a distance \dfrac{d}{2} (1 + \sqrt{3}) to the left of q .

Two point dipoles p\hat{k} and \dfrac{p}{2}\hat{k} are located at (0,0,0) and (1m,0,2m). Find the field at the point (1,0,0).

The given point is on axis of \displaystyle \frac{\bar{p}}{2} dipole and at equatorial line of \bar{p} dipole so that field at given point is \displaystyle \left ( \bar{E_{1}}+ \bar{E_{2}} \right ) \displaystyle \bar{E_{1}}=\frac{2K(p/2)}{2^{3}}=\frac{Kp}{8}(+\hat{k}) \displaystyle \bar{E_{2}}=\frac{Kp}{1}(-\hat{k}) \displaystyle \bar{E_{1}}+\bar{E_{2}}=\frac{7}{8}Kp(-\hat{k})=-\frac{7p}{32\pi \varepsilon _{0}}\hat{k}

The electric field components in figure are E_x=b\sqrt x, E_y=E_z=0, Calculate the net flux through the cube.

As E_y=E_z=0 so only faces along x-axis will contribute the flux through the cube.Entering flux through the face at x=a is \phi_{in}=E_x.a^2=b\sqrt{x}.a^2=b\sqrt{a}.a^2=ba^{5/2} Outgoing flux through the face at x=2a is \phi_{out}=E_x.a^2=b\sqrt{2a}.a^2=b\sqrt{2a}.a^2=ba^{5/2}\sqrt{2} Thus , net flux through the cube is \phi=\phi_{out}-\phi_{in}=ba^{5/2}\sqrt{2}-ba^{5/2}=ba^{5/2}(\sqrt 2-1)

Electric Charges And Fields - Medium level questions

Q: Two identical charged spheres are suspended by strings of equal length and the strings make certain angle with each other. When suspended in a liquid of density 400\ kg/m^{3} , the angle between the threads remains the same. If the density of the material of the sphere is 1600\ kg/m^{3} , the dielectric constant of the liquid:

Initially when the balls are in air let the angle be \theta between them \Rightarrow \dfrac{K q ^2}{4l^2sin^2(\frac{\theta }{2})}= T\ sin(\frac{\theta }{2})\ \ \ and\ T\ cos(\frac{\theta }{2})= mg \Rightarrow tan\dfrac{\theta }{2}= \dfrac{K q ^2}{4l^2sin^2(\frac{\theta }{2})mg} \Rightarrow tan \dfrac{\theta }{2} sin^2\dfrac{\theta }{2}= \dfrac{1}{4\pi \epsilon_o}\dfrac{q ^2}{4l^2\rho_bVg} Now when it is inserted in water we get electric K \varepsilon _0 and buoyant force = \rho_lVg \therefore \ Weight, \ \ \dfrac{1}{4\pi \epsilon_o}\dfrac{q ^2}{4l^2\rho_bVg}= \dfrac{q ^2}{4\pi (K\epsilon_o)4l^2(\rho_b-\rho_l)Vg} \Rightarrow K(\rho_b-\rho_l)= \rho_b \Rightarrow K= \dfrac{1600}{1600-400}= 1.33

Q: There is a uniform electric field of strength 10^{3}V/m along y-axis . A body of mass 1 g and charge 10^{-6}C is projected into the field from origin along the positive x-axis with a velocity 10 m/s . Its speed in m/s after 10s is (neglect gravitation) :

Velocity in x direction remains constant i.e, v_{x}=10\ m/s Now acceleration is only in y direction a=\dfrac{qE}{m} a=\dfrac{10^{3}\times 10^{-6}}{1\times 10^{-3}} \therefore Velocity in y direction v_{y}=a\times t v_{y}=10m/s \therefore v=10\hat{x}+10\hat{y} \left | v \right |=10\sqrt{2}

Q: An electric dipole is placed in an electric field generated by a point charge.

Hint: Find the net force on the dipole by using F=qE to calculate the force on both charges of the dipole.Explanation:Step 1: Reasons for incorrect options:For net electric force to be zero on the dipole, the electric field lines should be parallel to each other, and the dipole is placed in such a way that it is perpendicular to the electric field. But parallel field lines can be there as well for a point charge. So, options A and B are incorrect.Now, torque on the dipole is given by the formula: \vec{ \tau} = \vec{P} \times \vec{E} If the dipole moment P and electric field line E is in the same direction, in that case, torque will be zero in another case, it will have some non-zero value. So, option C is incorrect. Answer: Hence, option D is the correct answer.

Q: A charge of 10^{-10}C is palced at the origin. The eletric field(in N/C) at (1,1) cm due to it, is:

\vec{E}=\dfrac{k(10^{-10})}{(\sqrt{2}\times 10^{-2})^{2}}(\dfrac{\hat{x}}{\sqrt{2}}+\dfrac{\hat{y}}{\sqrt{2}}) =\dfrac{9\times 10^{9}\times 10^{-10}}{10^{-4}\times 2}\ \left ( {\dfrac{\hat{x}}{\sqrt{2}}+\dfrac{\hat{y}}{\sqrt{2}}} \right )\ \to unit vector in the above shown direction \vec{E}=\dfrac{4.5\times 10^{3}}{\sqrt{2}}\ (\hat{x}+\hat{y})

Q: A sphere of radius R carries charge such that its volume charge density is proportional to the square of the distance from the center. What is the ratio of the magnitude of the electric field at distance 2R from the center to the magnitude of the electric field at a distance of R/2 from the center?

Given that volume charge density \rho=Cr^2 Total charge on the sphere q=\int_{0}^{r}4 \pi r^2 dr\rho =\int_{0}^{r} 4 \pi Cr^4 dr =\cfrac{4}{5} \pi Cr^5 E_{r=R/2}\cfrac {kq_{(r=R/2)}}{(R/2)^2}=\cfrac{k(4/5)\pi C(R/2)^5}{ (R/2)^2} E_{r=2R}=\cfrac {kq_{(r=R)}}{(2R)^2}=\cfrac{k(4/5)\pi CR^5}{4 R^2} Now solve to get \cfrac{E_{r=2R}}{E_{r=R/2}}=2

Q: A cubical Gaussian surfaces encloses 30\ C per unit permittivity of charge. The electric flux through each face of the cube is

Given cube encloses \dfrac {30C}{\epsilon_o} we know gauss law \oint E\cdot da=\dfrac {q_o}{\epsilon_o} \therefore Total flux through all 6 faces =30 C per unit permittivity \Rightarrow Flux through each face =\dfrac {30}{6} \Rightarrow \dfrac {5C}{\epsilon_o}

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Medium level questions

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Electric Charges And Fields

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Two identical charged spheres are suspended by strings of equal length and the strings make certain angle with each other. When suspended in a liquid of density $400 k g / m^{3}$ , the angle between the threads remains the same. If the density of the material of the sphere is $1600 k g / m^{3}$ , the dielectric constant of the liquid:

1.33

2

3.12

5

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Two charges

q

and

- 3 q

are placed fixed on

x

-axis separated by distance

^{'} d^{'} .

Where should a third charge

2 q

be placed such that it will not experience any force?

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There is a uniform electric field of strength $1 0^{3} V / m$ along $y - a x i s$ . A body of mass $1 g$ and charge $1 0^{- 6} C$ is projected into the field from origin along the positive $x - a x i s$ with a velocity $10 m / s$ . Its speed in m/s after $10 s$ is (neglect gravitation) :

10

52

102

20

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An electric dipole is placed in an electric field generated by a point charge.

The net electric force on the dipole must be zero.

The net electric force on the dipole may be zero.

The torque on the dipole due to the field must be zero.

The torque on the dipole due to field may be zero.

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Two point dipoles

p \hat{k}

and

\frac{p}{2} \hat{k}

are located at (0,0,0) and (1m,0,2m). Find the field at the point (1,0,0).

\frac{9 p}{3 2 π ε _{0}} \hat{k}

\frac{- 7 p}{3 2 π ε _{0}} \hat{k}

\frac{7 p}{3 2 π ε _{0}} \hat{k}

none of these

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A charge of $1 0^{- 10} C$ is palced at the origin. The eletric field(in N/C) at $(1, 1)$ cm due to it, is:

(\overset{ˉ}{i} + \overset{ˉ}{j}) \times 1 0^{3}

\frac{4 . 5}{2} (\overset{ˉ}{i} + \overset{ˉ}{j}) \times 1 0^{3}

20 \overset{ˉ}{j} \times 1 0^{3}

4.52 (\overset{ˉ}{i} + \overset{ˉ}{j}) \times 1 0^{3}

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A hemisphere body of radius

R

is placed in a uniform electric field

E

. What is the flux linked with the curved surface if the field is perpendicular to the base?

2 π R^{2} E

π R^{2} E

π R^{2} E / 2

0

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The electric field components in figure are

E_{x} = b x, E_{y} = E_{z} = 0,

Calculate the net flux through the cube.

(2 - 1) b a^{\frac{5}{2}}

b a^{\frac{5}{2}}

b a^{\frac{5}{2}}

(5 - 3) b a^{\frac{5}{2}}

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A sphere of radius R carries charge such that its volume charge density is proportional to the square of the distance from the center. What is the ratio of the magnitude of the electric field at distance 2R from the center to the magnitude of the electric field at a distance of R/2 from the center?

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A cubical Gaussian surfaces encloses $30 C$ per unit permittivity of charge. The electric flux through each face of the cube is

30 C

15 C

10 C

5 C

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