Q2: The net magnetic flux through any closed surface is
(a) Zero
(b) Positive
(c) Infinity
(d) Negative
Ans: (a)
Magnetic monopole doesn't exist.
Hence net magnetic flux through any closed surface is zero.
Q1: Two point charges –q and +q are placed at a distance of L, as shown in the figure
The magnitude of electric field intensity at a distance R(R >>L) varies as:
(a) 1/R^{3}
^{(b) 1/R4}
(c) 1/R^{6}
(d) 1/R^{2}
Ans: (a)
For R >>L, arrangement is an electric dipole
Q1: Polar molecules are the molecules:
(a) acquire a dipole moment only when the magnetic field is absent.
(b) having a permanent electric dipole moment.
(c) having zero dipole moment.
(d) acquire a dipole moment only in the presence of an electric field due to displacement of charges.
Ans: (b)
Polar molecules have centers of positive and negative charges separated by some distance, so they have permanent dipole moments.
Q2: A dipole is placed in an electric field as shown. In which direction will it move?
(a) towards the left as its potential energy will decrease.
(b) towards the right as its potential energy will increase.
(c) towards the left as its potential energy will increase.
(d) towards the right as its potential energy will decrease.
Ans: (d)
E_{1} > E_{2} as field lines are closer at charge +q, so the net force on the dipole acts towards the right side.
A system always moves to decrease its potential energy.
Q1: A short electric dipole has a dipole moment of 16×10^{9} C m. The electric potential due to the dipole at a point at a distance of 0.6 m from the center of the dipole, situated on a line making an angle of 60° with the dipole axis is:
(a) 400 V
(b) Zero
(c) 50 V
(d) 200 V
Ans: D
Q3: A spherical conductor of radius 10 cm has a charge of 3.2 × 10^{−7} C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere?
(a) 1.28 × 10^{5} N/C
(b) 1.28 × 10^{6} N/C
(c) 1.28 × 10^{7} N/C
(d) 1.28 × 10^{4} N/C
Ans: (a)
Given, radius, r = 10 cm = 10 × 10 ^{−2 }m Charge, q = 3.2 × 10^{ −7} C Electric field, E = ?
Electric field at a point (x = 15 cm) from the centre of the sphere is
Hence, correct option is (a).
Q4: The electric field at a point on the equatorial plane at a distance r from the centre of a dipole having dipole moment r P is given by (r >> separation of two charges forming the dipole, ε 0 = permittivity of free space)
(a)
(b)
(c)
(d)
Ans: (a)
Electric field due to electric dipole on equatorial plane at a distance r from the centre of dipole is given as
Q1: A hollow metal sphere of radius R is uniformly charged. The electric field is due to the sphere at a distance r from the
center.
(a) Increases as r increases for both r < R and r > R
(b) Zero as r increases for r < R, decreases as r increases for r > R
(c) Zero as r increases for r < R, increases as r increases for r > R
(d) Decreases as r increases for both r < R and r > R
Ans: (b)
Charge Q will be distributed over the surface of a hollow metal sphere.
(i) For r < R (inside)
By Gauss law,
⇒ E_{in} = 0 (∵ q_{en} = 0)
(ii) For r > R (outside)
Q2: Two parallel infinite line charges with linear charge densities +λ C/m and λl C/m are placed at a distance of 2R in free space. What is the electric field midway between the two line charges?
(a) Zero
(b)
(c)
(d)
Ans: (c)
Electric field due to line charge (1)
Electric field due to line charge (2)
Q3: Two point charges A and B, having charges +Q and –Q respectively, are placed at a certain distance apart, and the force acting between them is F. If 25% charge of A is transferred to B, then the force between the charges becomes :
(a) F
(b) 9F/16
(c) 16F/9
(d) 4F/3
Ans: (b)
If 25% of the charge of A transferred to B then
Q1: An electron falls from rest through a vertical distance h in a uniform and vertically upwarddirected electric field E. The direction of the electrical field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest through the same vertical distance h. The time fall of the electron, in comparison to the time fall of the proton is:
(a) Smaller
(b) 5 times greater
(c) 10 times greater
(d) equal
Ans: (a)
Q2: A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric fieldDue to the forceits velocity increases from 0 to 6 m/s in onesecond duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively:
(a) 2 m/s, 4 m/s
(b) 1 m/s, 3 m/s
(c) 1 m/s, 3.5 m/s
(d) 1.5 m/s, 3 m/s
Ans: (b)
0 < t < 1s : velocity increases from 0 to 6 m/s
1 < t < 2s : velocity decreases from 6 to 0 m/s
but the car continues to move forward
2 < t < 3s : since field strength is same ⇒ same acceleration
∴ car's velocity increases
from 0 to –6 m/
Q1: Suppose the charge of a proton and an electron differ slightly. One of them is − e and the other is (e + ∆e). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then ∆e is of the order [Given mass of hydrogen, m_{h} = 1.67 × 10^{−27 }kg]
(a) 10^{−20} C
(b) 10^{−23} C
(c) 10 ^{−37}C
(d) 10^{ 47}C
Ans: (c)
Net charge on one Hatom q = − e + e + ∆e = ∆e Net electrostatic repulsive force between two Hatoms
Similarly, net gravitational attractive force between two Hatoms
Q1: Two identical charged spheres suspended from a common point by two massless strings of lengths ℓ are initially at a distance d (d << ℓ) apart because of their mutual repulsion. The charges begin to leak from both spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as:
(a) v ∝ x^{1}
(b) v ∝ x^{1/2}
(c) v ∝ x
(d) v ∝ x^{1/2}
Ans: (d)
Q2: An electric dipole is placed at an angle of 30° with an electric field of intensity 2 × 10^{5} N C^{1}. It experiences a torque equal to 4 N m. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.
(a) 8 mC
(b) 6 mC
(c) 4 mC
(d) 2 mC
Ans: (d)
E = 2 × 10^{5} N/C
l = 2 cm
τ = 4 Nm
4 = pEsinθ
4 = p × 2 × 10^{5} × sin30°
p = 4 × 10^{–5} cm
Q1: The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius = a centered at the origin of the field will be given by:
(a)
(b)
(c)
(d)
Ans: (d)
105 videos425 docs114 tests

1. What is the concept of electric charges and fields in physics? 
2. What is the SI unit of electric charge and how is it measured? 
3. How do positive and negative charges interact in an electric field? 
4. What is the formula for calculating the force between two charges in an electric field? 
5. How can electric fields be used in practical applications? 

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