The Gaussian surface for calculating the electric field due to a charge distribution is:

A symmetrical closed surface at every point of which electric field has a single fixed value.

Any closed surface around the charge distribution

Any surface near the charge distribution

Two copper spheres A and B of same size are charged to same potential. A is hollow and B is solid. Which of the two hold more charge?

Solid sphere cannot hold any change

Hollow sphere cannot hold any charge

If one penetrates a uniformly charged spherical shell, the electric field strength E:

Remains the same as at the surface

The figure is plot of lines of force due to two charges q1 and q2.; Figure out the sign of charges.

Left positive and right negative

Right negative and left positive

Electric Charges And Fields

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1/15 Questions

The Gaussian surface for calculating the electric field due to a charge distribution is:
- Any closed surface around the charge distribution
- Any surface near the charge distribution
- A spherical surface
- A symmetrical closed surface at every point of which electric field has a single fixed value.

About This Quiz

This quiz on Electric Charges and Fields assesses understanding of Gaussian surfaces, charge distribution, effects of electric fields on dipoles, and calculations involving forces between charges. It's designed to enhance learners' grasp of fundamental electromagnetism concepts, crucial for advanced studies in physics.

2.

Two copper spheres A and B of same size are charged to same potential. A is hollow and B is solid. Which of the two hold more charge?
- Solid sphere cannot hold any change
- Hollow sphere cannot hold any charge
- Both have zero charge
- Both have same charge
Correct Answer

A. Both have same charge

Explanation

The correct answer is that both spheres have the same charge. This is because the charge on an object depends on its potential and the size of the object, not on whether it is hollow or solid. Since both spheres are charged to the same potential and have the same size, they will hold the same amount of charge.

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2
3.

When an electric dipole is placed in a uniform electric field, it experiences
- A force as well as torque
- A torque but no force
- A force but no torque
- Neither a force nor a torque
Correct Answer

A. A torque but no force

Explanation

When an electric dipole is placed in a uniform electric field, it experiences a torque but no force. This is because a dipole consists of two opposite charges separated by a distance. The electric field exerts equal and opposite forces on the charges, resulting in a net force of zero. However, the electric field exerts a torque on the dipole, causing it to align itself with the field. The torque is given by the product of the dipole moment and the electric field strength. Therefore, the dipole experiences a torque but no net force.

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1
4.

Two point charges at a certain distance experience a force of 5 newton. Each charge is doubled in magnitude and distance between the two is halved. The interacting force would be:
- 16 N
- 80 N
- 5 N
- 20 N
Correct Answer

A. 80 N

Explanation

When the magnitude of each charge is doubled, the force between them increases by a factor of 4 (2^2). When the distance between the charges is halved, the force between them increases by a factor of 4 again (2^2). Therefore, the overall increase in force is 4 * 4 = 16 times the original force of 5 N. Hence, the interacting force would be 16 * 5 N = 80 N.

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5
5.

The S.I. unit for surface integral of electric field is:
- Vm
- V
Correct Answer

A. Vm

Explanation

The correct answer is Vm. The surface integral of electric field is a measure of the electric flux passing through a given surface. Electric flux is the product of electric field and the area vector of the surface. Since electric field is measured in volts per meter (V/m) and area is measured in square meters (m^2), the unit for surface integral of electric field is Vm.

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2
6.

The value of charge q at the centre of two equal and like charges Q so that the three are in equilibrium is:
- Q
- +Q/4
- –Q/4
- Q/2
Correct Answer

A. –Q/4

Explanation

When two equal and like charges Q are placed at the ends, the charge q at the center experiences a repulsive force from both charges. In order for the system to be in equilibrium, the net force on q must be zero. This can be achieved by choosing a charge q that is equal in magnitude but opposite in sign to Q/4. Therefore, the correct answer is –Q/4.

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6
7.

If one penetrates a uniformly charged spherical shell, the electric field strength E:
- Increases
- Decreases
- Remains the same as at the surface
- Is zero at all points
Correct Answer

A. Is zero at all points

Explanation

When one penetrates a uniformly charged spherical shell, the electric field strength E is zero at all points. This is because the electric field inside a uniformly charged spherical shell is zero. The charges on the shell are uniformly distributed on the outer surface, and due to the symmetry of the shell, the electric field vectors from each charge cancel each other out at all points inside the shell. Therefore, the electric field strength inside the shell is zero.

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1
8.

Electric field at the centroid of a triangle carrying q charge at each corner is:
- Zero
Correct Answer

A. Zero

Explanation

The electric field at the centroid of a triangle carrying equal q charges at each corner is zero because the electric fields produced by the charges cancel each other out due to the symmetry of the triangle. The forces exerted by the charges on the centroid are equal in magnitude and opposite in direction, resulting in a net electric field of zero.

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1
9.

There is an electric field E in X-direction. If work done in moving a charge 0.2 c through a distance of 2m along a line making an angle of 60 degree with X-axis is 0.4 joule, what is the value of E?
- 4 N / C
- 5 N / C
- None of these
Correct Answer

A. None of these
10.

A charged spherical shell does not produce an electric field at any:
- Interior point
- Outer point
- Beyond 2 meters
- Beyond 10 meters
Correct Answer

A. Interior point

Explanation

A charged spherical shell does not produce an electric field at any interior point because the electric field inside a conductor is always zero. This is because the charges in the conductor rearrange themselves in such a way that the electric field inside cancels out. Therefore, any point inside the charged spherical shell will not experience an electric field.

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11.

A positively charged ball hangs from a silk thread. We put positive test charge q₀ at a point and measure F/q₀ then it can be predicted that the electric field strength E is:
- > F/q₀
- = F/q₀
- < F/q₀
- None of the above
Correct Answer

A. > F/q₀

Explanation

The correct answer is "> F/q0". This implies that the electric field strength E is greater than the force per unit charge F/q0. This can be understood by considering the fact that the positively charged ball is attracting the positive test charge q0. The force experienced by the test charge is directly proportional to the electric field strength, and since the force is greater than the charge, the electric field strength must also be greater.

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12.

The figure is plot of lines of force due to two charges q₁ and q₂.; Figure out the sign of charges.
- Both negative
- Both positive
- Left positive and right negative
- Right negative and left positive
Correct Answer

A. Both positive

Explanation

The correct answer is "Both positive". This can be determined by observing the direction of the lines of force in the plot. Since like charges repel each other, the fact that the lines of force are moving away from both charges indicates that they are both positive.

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1
13.

If the electric flux entering and leaving an enclosed surface respectively is and , the electric charge inside the surface will be:
Correct Answer

A.
14.

Two equal metal balls are charged to 10 and -20 units of electricity. Then they are brought in contact with each other and then again separated to original distance. The ratio of the magnitudes of forces between the balls before and after contact is:
- 8 : 1
- 1 : 8
- 1 : 2
- 2 : 1
Correct Answer

A. 8 : 1
15.

The unit permittivity of free space is:
- C/N-m
Correct Answer

A.

Explanation

The unit permittivity of free space is C/N-m. This unit represents the capacitance per unit length in a vacuum. It is a measure of how easily electric field lines can pass through a vacuum. The higher the permittivity, the more electric field lines can pass through, indicating a higher ability to store electric charge. The value of the unit permittivity of free space is approximately 8.854 x 10^-12 C/N-m.

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1

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Apr 28, 2023

Quiz Edited by
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Nov 21, 2013

Quiz Created by
Tanmay Shankar

Electric Charges And Fields

The Gaussian surface for calculating the electric field due to a charge distribution is:

Quiz Preview

Two copper spheres A and B of same size are charged to same potential. A is hollow and B is solid. Which of the two hold more charge?

When an electric dipole is placed in a uniform electric field, it experiences

Two point charges at a certain distance experience a force of 5 newton. Each charge is doubled in magnitude and distance between the two is halved. The interacting force would be:

The S.I. unit for surface integral of electric field is:

The value of charge q at the centre of two equal and like charges Q so that the three are in equilibrium is:

If one penetrates a uniformly charged spherical shell, the electric field strength E:

Electric field at the centroid of a triangle carrying q charge at each corner is:

There is an electric field E in X-direction. If work done in moving a charge 0.2 c through a distance of 2m along a line making an angle of 60 degree with X-axis is 0.4 joule, what is the value of E?

A charged spherical shell does not produce an electric field at any:

A positively charged ball hangs from a silk thread. We put positive test charge q0 at a point and measure F/q0 then it can be predicted that the electric field strength E is:

The figure is plot of lines of force due to two charges q1 and q2.; Figure out the sign of charges.

If the electric flux entering and leaving an enclosed surface respectively is and , the electric charge inside the surface will be:

Two equal metal balls are charged to 10 and -20 units of electricity. Then they are brought in contact with each other and then again separated to original distance. The ratio of the magnitudes of forces between the balls before and after contact is:

The unit permittivity of free space is:

A positively charged ball hangs from a silk thread. We put positive test charge q₀ at a point and measure F/q₀ then it can be predicted that the electric field strength E is:

The figure is plot of lines of force due to two charges q₁ and q₂.; Figure out the sign of charges.