The **surface Charge density** of a conductor refers to the amount of **electric charge** distributed per unit area on the surface of the conductor. It is one of the important topics in Electrostatics. Because the charge densities are used to determine the **electric fields** due to different distributions of charge on the conductors. In this article, I’m going to reveal **7 most important MCQ questions on Surface Charge Density**. Answers to these questions are also given here with proper explanations.

## Theoretical MCQ questions on Surface Charge Density

**1)** **What is the Dimensional formula of Surface Charge Density?**

a) [LTI]

b) [L^{-1}TI]

c) [L^{-2} TI]

d) [LTI^{-1}]

**Answer: Option – (c)**.

**Explanation:**

Formula for surface Charge density = \color{Blue}\frac{Charge}{Area}. The electric charge has the dimension of [**TI**] and the area has a dimension of [**L ^{2}**]. Hence the dimension of the surface Charge density is \color{Blue}\frac{[TI]}{[L^{2}]}=[L^{-2}TI].

**2. How does the surface charge density depend on the curvature of the surface?**

a) Surface charge density is lower if the curvature is greater.

b) Surface charge density is greater if the curvature is greater.

c) Follows inverse square law of curvature.

d) Independent of curvature.

**Answer: Option – (b)**.

**Explanation:**

Take the example of a sphere. The surface charge density is \color{Blue}\sigma = \frac{Q}{4\pi r^2}. The greater **curvature** means a lower radius and the equation shows that for a lower radius the density of charge is greater.

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

a) A conducting Rod.

b) A Spherical Conductor.

c) A conducting Cone.

d) A cylindrical conductor.

**Answer:** **Option – (b)**.

**Explanation:**

The surface charge density of a spherical conductor of charge **Q** and radius **r** is \color{Blue}\sigma = \frac{Q}{4\pi r^2}. Since the radius is the same from every point of its surface (due to the same curvature), the surface charge density is also the same at every elementary portion of its surface.

**4. Which of the followings is not a unit of surface charge density?**

a) C.m^{-2}

b) StatC.cm^{-2}

c) A.s.m^{-2}

d) A.s^{-1}.m^{-2}

**Answer: Option – (d)**.

**Explanation:**

We know that \color{Blue}\sigma = \frac{Charge}{Area}. Then SI and CGC units of electric charge are **Coulomb** **(C)** and **StatC** respectively and those for the surface area are **m ^{2}** and

**cm**respectively. Therefore, the SI unit of surface charge density is

^{2}**C.m**and the CGS unit is

^{-2}**StatC.cm**. Again, electric charge = (

^{-2}**electric current**

*×*time) which has a unit of

**Ampere.Second**(

**A.s**). Then A.s.m

^{-2}is another valid unit. But A.s

^{-1}.m

^{-2}is not a unit of \sigma.

**5. If a spherical shell of radius r contains total charge Q on its surface, then what is the surface charge density on its inner surface?**

a) \sigma = \frac{Q}{4\pi r^2}

b) \sigma = \frac{Q}{4\pi r}

c) \sigma = \frac{Q}{2\pi r}

d) Zero

**Answer: Option – (d)**.

**Explanation:**

Spherical shell is a hollow conductor. Electric charge lies on the outer surface of a hollow conductor. There will be no charge on the inner surface of the shell and hence the surface charge density will be zero for it.

## Numerical Problems on Surface Charge density

**6. Two Coulomb charges are distributed uniformly on the surface of a sphere of diameter 2 cm. The density of charge on its surface is**

a) 7.96 C.m^{-2}

b) 15.92 C.m^{-2}

c) 3.98 C.m^{-2}

d) 0.16 C.m^{-2}

**Answer:** **Option – (b)**.

**Explanation:**

The surface charge density of a sphere of charge **Q** and radius **r** is \color{Blue}\sigma = \frac{Q}{4\pi r^2}. Here, Q = 2 C and r = 1 cm = 0.01 m. Putting these values in this equation we get \sigma = 15.92 C.m^{-2}.

**7. Q amount of charge is distributed uniformly on the surface of a solid cylinder of radius r and length 2r. The ratio of the surface charge densities of the curved surface to each flat surface is**

a) 1 : 4

b) 4 : 1

c) 1 : 2

d) 2 : 1

**Answer:** **Option – (a)**.

**Explanation:**

The surface area of the curved surface and each flat surface of a cylinder are \color{Blue}\sigma = \frac{Q}{2\pi r L} and \color{Blue}\sigma = \frac{Q}{\pi r^2} respectively. Here length, L = 2r. Now, if you do the ratio you will get 1 : 4.

These are the **most important MCQs on surface charge density in electrostatics**. Hope you liked the questions and explanations of the answers. Please share the link of this article with your classmates who are looking for such kinds of questions.

Thank you.

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