We have already learned **electric charges** and **electric field** in detail. Now, it’s time to practice some multiple-choice questions on **electric charges and electric fields for class 12 physics**. It is very helpful for board exams and NEET exam. This topic is included in class 12 NCERT physics chapter 1 for CBSE and other boards. So, in this article, we are going to discuss **MCQ questions on electric field for class 12 and class 10**. If you need a PDF file, send me an email.

**Class 12 Physics**

**Chapter 1 : Electric charges and fields**

**MCQ on electric field**

#### 1. **An object has total charge of -80 micro-coulomb on it. The number of excess electrons that the object has is**

- 5×10
^{10} **5×10**^{14}- 1.6×10
^{12} - 1.6×10
^{10}

**2. Which of the followings is not the unit of electric charge?**

- Coulomb
- esu
- A.s
**A.m**

**3. The dimensional formula of electric field intensity (E) is**

**MLT**^{-3}I^{-1}- MLT
^{-3}I^{-2} - LTI
^{-3} - ML
^{2}T^{-3}I^{-1}

**4. Which of the following conductors has equal charge density at every point on its surface?**

- Cylindrical wire
**Spherical conductor**- Cubic conductor
- rectangular conductor

See the article on **surface charge density of a conductor**.

**4. Which physical quantity has the unit of V.m**^{-1}?

^{-1}?

- Electric potential
- Charge
**Electric field intensity**- Electric Flux

**5. What is the unit of electric flux?**

- Volt
- N/C
**V.m**- V.m
^{-1}

**6. The dimensional formula of the electric flux is**

**ML**^{3}T^{-3}I^{-1}- ML
^{2}T^{-3}I^{-1} - L
^{2}T^{-3}I - ML
^{2}T^{-3}I

**7. A sphere of radius 10 cm has total charge q inside it. Then the electric flux through the spherical surface is**

- Zero
- \small \color{Blue}\frac{q}{\epsilon _{0}}
- \small \frac{q}{10}
- 10\small \epsilon

We can get this from **Gauss’s law of electrostatics.**

**8. What is the dimension of surface charge density?**

- ML
^{2}T^{-3}I^{-1} - L
^{2}T^{-3}I **L**^{-2}TI- ML
^{2}T^{-3}I

**9. The electric field intensity inside a hollow sphere of charge Q and radius R is**

**Zero**- \small \frac{Q}{\epsilon _{0}}
- \small \frac{Q}{4\pi \epsilon _{0}R^{2}}
- cannot be determined without knowing the distance from the center.

There is no charge inside a hollow sphere or a **spherical shell**. So, the electric field inside a hollow sphere is always zero.

**10. The electric field at any point** **due to an electric dipole varies with the distance r as**

- \small \frac{1}{r}
- r
^{2} - \small \frac{1}{r^{2}}
- \small \color{Blue} \frac{1}{r^{3}}

**11. The electric field at an axial point of a dipole is E**_{1} and at the same distance above the bisector of the dipole is E_{2}. Then the ratio E_{1} : E_{2} is

_{1}and at the same distance above the bisector of the dipole is E

_{2}. Then the ratio E

_{1}: E

_{2}is

- 1 : 1
**2 : 1**- 1 : 2
- 3 : 1

**12. Which of the followings is not a property of electric field lines?**

- Electric field lines start from positive charge and end at a negative charge.
- The tangent at any point on an electric field line gives the direction of the electric field at that point.
**Electric field lines are closed lines**.- A free positive charge will move along the electric field line if it is placed in an electric field.

Electric field lines are not closed. There exist a single positive or negative charge. So, electric field lines starting from a charge can be extended up to infinity. But the magnetic field lines are closed curves as the magnetic monopole does not exist. This is the difference between electric field lines and magnetic field lines. For further information, you can visit these articles – **Properties of electric field lines** and **properties of magnetic field lines**.

**13. Which of the following amount of charges cannot be produced in a body?**

- 3 C
- 4 C
**4×10**^{-19}C- 4.8×10
^{-19}C

**14. An electric charge q is placed at the center of a cube. The electric flux through each surface of the cube is**

- \small \frac{q}{\epsilon _{0}}
- Zero
- \small \color{Blue}\frac{q}{6\epsilon _{0}}
- \small \frac{q}{24\epsilon _{0}}

**15. An electric field of 10 V/m is present along the positive X-axis. Then the electric flux linked with the rectangular surface of area 10 m**^{2} placed in X-Y plane is

^{2}placed in X-Y plane is

**zero**- 100 V.m
- 1 V.m
- 10 V.m

Electric flux = \small {\color{Blue} \vec{E}.\vec{A}}. Here, electric field E is along X-axis and area A is along Z-axis as the plane is in X-Y plane. So, E and A are perpendicular to each other. Therefore, the electric flux through the surface becomes zero.

**16. An electric charge q is placed at r distance from a square plate of area A. The electric flux through the plate is**

- \small \color{Blue}\frac{q}{6\epsilon _{0}}
- Zero
- \small \frac{q}{\epsilon _{0}}
- qA/r

**17. The magnitude of electric force between two charges +2Q and -Q is F .** **Then they are brought in contact and separated again by the same distance.** **If the new force between them is F’ then the ratio F : F’ is**

- 1 : 2
- 4 : 1
**8 : 1**- 1 : 4

**18. A charge Q is enclosed by a Gaussian surface of radius r. If the radius is doubled, then the electric flux through the surface becomes**

- Double.
- Half.
**remains same**.- Four times.

Electric flux is independent of the shape of the conductor. It only depends on the amount of electric charge inside the conductor.

This is all from the article **MCQ on electric field and charge**. This article will be helpful for the students of class 12 and 10 of CBSE board and all other boards. If you have any doubt on this topic or if you have any other question on this topic you can ask me in the comment section.

Thank you!

**Related posts:**

**MCQ on electric potential****MCQ on current electricity****Gauss’s law of electrostatics****Electric field and electric field intensity****Electric potential and potential energy****Capacitor and capacitance**

In last second question ,how come the answer is 8:1 ???

Find the forces in both cases. In first case, charges are 2Q and Q. In the second case, charges will be Q/2 on each body. Distance is same in both cases. Then find the ratio of the forces. You will get the result.