This set of Class 12 Physics Chapter 1 Multiple Choice Questions & Answers (MCQs) focuses on “Electric Field”.
1. The amount of force exerted on a unit positive charge in an electric field is known as _____
a) Electric field intensity
b) Electric flux
c) Electric potential
d) Electric lines of force
View Answer
Explanation: The zone near a charge where its attraction or repulsion force works, is known as the electric field of that charge. Theoretically, it is up to infinite but practically it has limitations. If a unit positive charge is kept in that field, it will undergo some force which is known as electric field intensity at that point.
2. The direction of electric field created by a negative charge is ___________
a) Directed outwards
b) Directed towards the charge
c) Maybe outwards or towards the charge
d) Circular in shape
View Answer
Explanation: If a unit positive charge is kept near a negative charge, the unit positive charge will be attracted towards the negative charge. That means the electric field is towards the negative charge. But in case of positive charge, the field is directed away from the charge.
3. Electric field inside a hollow conducting sphere ______
a) Increases with distance from the center of the sphere
b) Decreases with distance from the center of the sphere
c) Is zero
d) May increase or decrease with distance from the center
View Answer
Explanation: According to Gauss’s law, if there is no charge inside a closed surface, the field inside the closed surface will always be zero. We know the charge is distributed on the outer surface of a conducting hollow sphere because the charges want to maintain maximum distance among them due to repulsion. So there is no charge inside the sphere and hence no electric field.
4. Electric field due to a uniformly charged hollow sphere at a distance of r (where r is greater than the radius of the sphere) is __________
a) Proportional to r
b) Inversely proportional to r
c) Proportional to r2
d) Inversely proportional to r2
View Answer
Explanation: If the total charge of the sphere is q then the electric field at a distance of r is equal to \(\frac {q}{4\pi\varepsilon_o r^2}\).Therefore the electric field is proportional is \(\frac {1}{r^2}\) (if r > radius of the sphere). But if r < radius of the sphere the electric field will be zero i.e. electric field inside a hollow sphere is always zero.
5. Two point charges q1 and q2 are situated at a distance d. There is no such point in between them where the electric field is zero. What can we deduce?
a) There is no such point
b) The charges are of the same polarity
c) The charges are of opposite polarity
d) The charges must be unequal
View Answer
Explanation: If both the charges are of the same polarity (maybe of unequal magnitude), there must be a point in between them where the electric field intensities of the charges are of equal magnitude and in opposite direction. Hence they balance each other and the net field intensity must be zero. But if the charges are of opposite polarities their field intensities aid each other and net field intensity can never be zero.
6. A uniformly charged sphere of radius R has charge +Q. A point charge –q is placed at a distance of 2R from the center of the sphere. The point charge will execute the simple harmonic motion. The statement is _____
a) False
b) True
View Answer
Explanation: In the case of SHM, the force on a body is inversely proportional to the distance of the body from the mean position. But in this case, we know the force acting on the body is inversely proportional to the square of the distance [F =\(\frac {qQ}{4\pi\varepsilon_o(2R)^2}\)]. So the motion of the body will be oscillatory but not SHM.
7. Two point charges of the same polarities are hung with the help of two threads and kept close. The angle between the threads will be _________ if the system is taken to space.
a) 180 degree
b) 90 degree
c) 45 degree
d) 60 degree
View Answer
Explanation: There is gravitational field on earth, so if we hang the two same charges there will be an interaction of vertical gravitational field and horizontal electric field. The system will achieve equilibrium by creating a certain angle between the threads and hence the vertical and horizontal components of forces will balance. But in space, there is no gravity. So the charges will be at 180-degree separation.
8. An electron of mass m is kept in a vertical electric field of magnitude E. What must be the value of E so that the electron doesn’t fall due to gravity?
a) m*g*e
b) \(\frac {e}{(m*g)}\)
c) \(\frac {(m*g)}{e}\)
d) \(\frac {1}{(m*g*e)}\)
View Answer
Explanation: Gravitational force on the electron is m*g (weight of the electron). Electrical force on the body is e*E. If the electron doesn’t fall then these two forces balance each other, so m*g=E*e. Therefore E= \(\frac {(m*g)}{e}\).
9. Electric field is a _______
a) Scalar quantity
b) Vector quantity
c) Tensor quantity
d) Quantity that has properties of both scalar and vector
View Answer
Explanation: A scalar quantity is a quantity with magnitude only but no direction. But a vector quantity possesses both magnitude and direction. An electric field has a very specific direction (away from a positive charge or towards a negative charge). Hence electric field is a vector quantity. Moreover, we have to use a vector addition for adding two electric fields.
10. Two point charges +4q and +q are kept at a distance of 30 cm from each other. At which point between them, the field intensity will be equal to zero?
a) 15cm away from the +4q charge
b) 20cm away from the +4q charge
c) 7.5cm away from the +q charge
d) 5cm away from the +q charge
View Answer
Explanation: The electric field at a distance of r from a charge q is equal to \(\frac {q}{4\pi\varepsilon_or^2}\). Let the electric field intensity will be zero at a distance of x cm from +4q charge, so the fields due to the two charges will balance each other at that point. Therefore \(\frac {4q}{4\pi\varepsilon_ox^2}=\frac {q}{4\pi\varepsilon_o(30-x)^2}\). Solving this we get x=20cm. Therefore the point will be 20cm away from the +4q charge.
11. What is the dimension of electric field intensity?
a) [M L T-2 I-1]
b) [M L T-3 I-1]
c) [M L T-2 I-2]
d) [M L T-3 I]
View Answer
Explanation: Electric field intensity is defined as the force on a unit positive charge kept in an electric field. Hence we can simply consider its dimension as\(\frac {the \, dimension \, of \, force}{the \, dimension \, of \, charge}\). The dimension of force is [MLT-2] and the dimension of charge is [IT]. Therefore the dimension of field intensity is [M L T-3 I-1].
12. V/m is the unit of ______
a) Electric field intensity
b) Electric flux
c) Electric potential
d) Charge
View Answer
Explanation: E=-\(\frac {dV}{dx}\) where E is the field intensity, V is potential and x is distance. Therefore unit of electric field intensity will be \(\frac {unit \, of \, potential}{unit \, of \, distance} = \frac {V}{m}\). Electric flux has unit V*m, V is the unit of electric potential whereas charge has a unit of Coulomb or esu.
13. Electric field intensity at the center of a square is _____ if +20 esu charges are placed at each corner of the square having side-length as 10 cm.
a) 0
b) 0.4 dyne/esu
c) 2 dyne/esu
d) 1.6 dyne/esu
View Answer
Explanation: Distance of center from each corner point of the square is = \(\frac {10\sqrt2}{2}\) = 5√2. Therefore field intensity at the center due to a single charge is = \(\frac {20}{(5\sqrt2)^2}\) dyne/esu. But the fields due to the four charges are equal and are at perpendicular to each other. So the fields balance each other and the net electric field at the center will be equal to zero.
14. A ball of 80mg mass and a 2*10-8 charge is hung with a thread in a uniform horizontal electric field of 2*104V/m. What is the angle made by the thread with vertical?
a) 27 degree
b) 30 degree
c) 45 degree
d) 0 degree
View Answer
Explanation: According to the diagram, Tsinθ = Eq; Tcosθ = mg. Where T is the tension in the thread, m is the mass of the ball, E is the electric field and q is the charge of the ball. Dividing these two, we get tanθ = \(\frac {Eq}{mg}\). Now substituting the value of E=2*104V/m, m=80mg, q=2*10-8C, g=9.8m/s2, we get tanθ = \(\frac {25}{49}\). Therefore θ=27 degree.
15. Find the electric field intensity at 10cm away from a point charge of 100 esu.
a) 1 dyne/esu
b) 10 dyne/esu
c) 100 dyne/esu
d) 9*109 dyne/esu
View Answer
Explanation: We know that electric field intensity at r distance from a point charge q is defined as \(\frac {q}{r^2}\) in the CGS system. Here q=100 esu, r=10cm. Substituting the value we get E=1dyne/esu. We must remember that factor \(\frac {1}{4\pi\varepsilon}\) is considered only in the SI system and its value is 9*109, but in the CGS system, we simply take the value of this constant as unity.
Sanfoundry Global Education & Learning Series – Physics – Class 12.
To practice all chapters and topics of class 12 Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
- Practice Class 12 - Biology MCQs
- Check Class 12 - Physics Books
- Practice Class 11 - Physics MCQs
- Practice Class 12 - Mathematics MCQs
- Practice Class 12 - Chemistry MCQs
