Electric Charges And Fields

1. 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

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.

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.

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

Interior point

Outer point

Beyond 2 meters

Beyond 10 meters

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.

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.

3. 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

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.

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.

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

Zero

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.

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.

5. 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

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.

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.

6. 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.

The correct answer is a symmetrical closed surface at every point of which electric field has a single fixed value. This is because a Gaussian surface is a hypothetical surface used to simplify calculations of electric fields. It is chosen such that the electric field is constant and symmetric at every point on the surface. This allows for easy integration of the electric field over the surface to calculate the total flux through the surface. A symmetrical closed surface ensures that the electric field has a single fixed value at every point, making the calculations more manageable.

Explanation

The correct answer is a symmetrical closed surface at every point of which electric field has a single fixed value. This is because a Gaussian surface is a hypothetical surface used to simplify calculations of electric fields. It is chosen such that the electric field is constant and symmetric at every point on the surface. This allows for easy integration of the electric field over the surface to calculate the total flux through the surface. A symmetrical closed surface ensures that the electric field has a single fixed value at every point, making the calculations more manageable.

7. 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

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.

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.

8. If the electric flux entering and leaving an enclosed surface respectively is

and

, the electric charge inside the surface will be:

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Explanation

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

Vm

V

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.

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.

10. 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

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.

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.

11. 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

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Explanation

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12. 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

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.

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.

13. 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

Explanation

14. 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

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.

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.

15. The unit permittivity of free space

is:

C/N-m

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.

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.