If the electric flux entering and leaving an enclosed surface respectively are \phi_1 and \phi_2 the electric charge inside the surface will be:

Net electric flux , \phi=\phi_2-\phi_1 Applying Gauss's law, \phi=\int \vec E.\hat{n}ds=\dfrac{Q_{enclosed}}{\epsilon_0} So, Q_{enclosed}=\phi \epsilon_0=(\phi_2-\phi_1)\epsilon_0

Electric Charges And Fields

Q: Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at the centre O, force on it will be:

The charges will be balanced by their counterparts on opposite side. So eventually the charges remaining will be 2q on B and 3q on D.Since the charge distribution is unsymmetrical, the net force on charge would be skewed towards D.

Q: 4 charges are placed each at a distance 'a' from origin. The dipole moment of configuration is :

As the total charge of the configuration is zero so the origin is independent of choice. Now we can find the dipole moment about given origin. Thus, dipole moment about the origin is \vec p=(-2q)(a\hat i)+3q(a\hat j)+(-2q)(-a\hat i)+q(-a\hat j)=2qa\hat j

Q: A and B are two points on the axis and the perpendicular bisector respectively of a electric dipole. A and B are far away from the dipole and at equal distance from it. The field at A and B are :

The electric field intensity due to the dipole at the point A on the axis of the dipole is given by E_A = \dfrac {2kp}{r^3} ......(1)where, p is electric dipole moment, r is the distance of point A from dipole and k = \dfrac{1}{4 \pi \epsilon_0} The direction of the electric field E is along the axis of the dipole from the negative charge towards the positive charge.Also, The electric field intensity due to the dipole at the point B on the perpendicular bisector of the dipole is given by E_B = \dfrac{kp}{r^3} ......(2)The direction of the electric field E is parallel to the axis of the dipole from the positive charge towards the negative charge.Hence, direction of electric field intensity due to the dipole at the point A is opposite to that of at point B.(refer the figure)From (1) and (2), we get \vec{E_A} = - 2 \vec{E_B}

Q: An electric dipole is placed in a uniform electric field. The net electric force on the dipole

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: Concept used:Dipole consists of 2 equal and opposite charges +q \ and -q . Consider a dipole, as shown below in the figure. In uniform electric field E, if we place any dipole at that time, Step 2: Calculation of force:Force on +q is q E and force on charge -q is -q E . So, the total force will be sum of these two forces, and it will result in zero. Answer: Hence, option A is the correct answer.

Q: Two large parallel planes charged uniformly with surface charge densities \sigma and -\sigma are located as shown in the figure. Which one of the following graphs shows the variation of electric field along a line perpendicular to the planes as one moves from A to B?

Electric field by a plane of charge density \sigma is given by \dfrac {\sigma}{2\epsilon_0} Thus,In the region I and III net electric field is zero.In the region II net electric field is \dfrac {\sigma}{2\epsilon_0}+\dfrac {\sigma}{2\epsilon_0}=\dfrac {\sigma}{\epsilon_0} and is uniform.

Q: When a negative charge is released and moves in electric field, it moves toward a position of :

Hint:The direction of the Electric Field is from the Positive end to the Negative end.Correct Option is C.Explanation for the correct answer: \bullet When a negative charge moves in an electric field, it gets attracted by the Coulomb force towards the positive end. Since the positive end is at higher potential, the negative charge moves towards a position of higher potential when introduced in an Electric Field. options A and B are incorrect. \bullet Since the force between the negative charge and the positive end is attractive in nature, the Potential Energy of the system is negative and it is more stable. Thus, the negative charge moves towards a position of lower Potential Energy. Option D is incorrect.Thus, Option C is correct.

Q: A closed surface in the shape of a cube is placed in a uniform electric field E along one of the sides of the cube as shown here. The length of the side is l . The total electric flux through two faces 1 and 2 is:

Electric flux within any closed surface in region of uniform electric field is zero because the total number of electric lines of forces entering the closed surface equals that exiting the surface. Hence , total flux, \phi_{tot}=\phi_{in}(2)+\phi_{out}(1) =El^2-El^2=0

Q: A cylinder of radius R and length L is placed in the uniform electric field E parallel to the cylinder axis. The total flux from the curved surface of the cylinder is given by :

As shown in the diagram angle between the curved surface and the field E is 90^0 We know than flux =E.da flux=E.da cos 90^0 flux=0 \therefore flux\ from \ curved\ surface =0

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Easy Questions

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Electric Charges And Fields

- It is always better to start confident. Let's practice easy questions to serve the purpose

Two identical pendulums

A

and

B

are suspended from the same point. Both are given a positive charge, with

A

having more charge than

B

. They diverge and reach equilibrium with the strings of pendulums

A

and

B

making angles

θ_{1}

and

θ_{2}

with the vertical respectively. Then:

θ_{1}

θ_{2}

θ_{1}

θ_{2}

θ_{1} = θ_{2}

the tension in A is greater than that in B

View solution

Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at the centre O, force on it will be:

zero

along OF

along OC

none of these

View solution

4

charges are placed each at a distance 'a' from origin. The dipole moment of configuration is :

2 q a j

3 q a j

2 a q [i + j]

none

View solution

A

and

B

are two points on the axis and the perpendicular bisector respectively of a electric dipole.

A

and

B

are far away from the dipole and at equal distance from it. The field at

A

and

B

are :

E_{A} = E_{B}

E_{A} = 2 E_{B}

E_{A} = - 2 E_{B}

∣ E_{B} ∣ = \frac{1}{2} ∣ E_{A} ∣, a n d E_{B} i s p e r p e n d i c u l a r t o E_{A}

View solution

An electric dipole is placed in a uniform electric field. The net electric force on the dipole

is always zero

depends on the orientation of the dipole

depends on the dipole moment

is always finite but not zero

View solution

Two large parallel planes charged uniformly with surface charge densities

σ

and

- σ

are located as shown in the figure. Which one of the following graphs shows the variation of electric field along a line perpendicular to the planes as one moves from A to B?

View solution

When a negative charge is released and moves in electric field, it moves toward a position of :

lower electric potential and lower potential energy

lower electric potential and higher potential energy

higher electric potential and lower potential energy

higher electric potential and higher potential energy

View solution

A closed surface in the shape of a cube is placed in a uniform electric field

E

along one of the sides of the cube as shown here. The length of the side is

l

. The total electric flux through two faces 1 and 2 is:

zero

- E l^{2}

2 E l^{2}

\frac{E l ^{2}}{2}

View solution

If the electric flux entering and leaving an enclosed surface respectively are

ϕ_{1}

and

ϕ_{2}

the electric charge inside the surface will be:

(ϕ_{1} + ϕ_{2}) ϵ_{0}

(ϕ_{2} - ϕ_{1}) ϵ_{0}

\frac{ϕ _{1} + ϕ _{2}}{ϵ _{0}}

\frac{ϕ _{2} - ϕ _{1}}{ϵ _{0}}

View solution

A cylinder of radius R and length L is placed in the uniform electric field E parallel to the cylinder axis. The total flux from the curved surface of the cylinder is given by :

2 π R^{2} E

\frac{π R ^{2}}{E}

\frac{π R ^{2} - π R}{E}

z e r o

View solution