IMPORTANT QUESTIONS CHAPTER - 1 : ELECTRIC CHARGES AND FIELDS

CHAPTER – 1 : ELECTRIC CHARGES AND FIELDS

IMPORTANT QUESTIONS

CHAPTER – 1 : ELECTRIC CHARGES AND FIELDS

1. How much positive and negative charge is there in a cup of water (250 gram)?

Sol. The molecular mass of water is 18g. Thus, one mole (= 6.02 × 10²³ molecules) of water is 18 g. Therefore the number of molecules in one cup of water is (250/18) × 6.02 × 10²³. Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons.

Hence the total positive and total negative charge has the same magnitude.

It is equal to (250/18) × 6.02 × 10²³ × 10 × 1.6 × 10^–19 C = 1.34 × 10⁷ C.

2. Four point charges q_A = 2μC, q_B = –5μC, q_C = 2μC, and q_D = –5μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1μC placed at the centre of the square?

Sol.

$\displaystyle {{F}_{{OA}}}=\frac{{k\left( {2\mu C} \right)\left( {1\mu C} \right)}}{{A{{O}^{2}}}}$

$\displaystyle {{F}_{{OB}}}=\frac{{k\left( {5\mu C} \right)\left( {1\mu C} \right)}}{{B{{O}^{2}}}}$

$\displaystyle {{F}_{{OC}}}=\frac{{k\left( {2\mu C} \right)\left( {1\mu C} \right)}}{{C{{O}^{2}}}}$

$\displaystyle {{F}_{{OD}}}=\frac{{k\left( {5\mu C} \right)\left( {1\mu C} \right)}}{{D{{O}^{2}}}}$

In a square AO = BO = CO = DO

So F_OA = F_OC

F_OB = F_OD

These force are same in magnitude but opposite in direction so resultant force will be equal to

zero. F_R = 0.

3. A system has two charges q_A = 2.5 × 10^–7C and q_B = – 2.5 × 10^–7C located at points A : (0, 0, –15 cm) and B : (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

Sol. Total charge = 2.5 × 10^–7 – 2.5 × 10^–7 = 0

Electric dipole moment is $\displaystyle \overrightarrow{p}=q\left( {2\overrightarrow{a}} \right)$

= 2.5 × 10^–7 × (0.15 + 0.15) C–m

= 7.5 × 10^–8 C–m

The direction of dipole moment is along –Z Axis.

4. Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole of dipole moment and length 2a. What is the direction of this field ?

Sol.

We consider a dipole consisting of –q and +q separated by a distance 2a. Let P be a point on the

equatorial line.

$\displaystyle {{\overrightarrow{E}}_{A}}=\frac{1}{{4\pi {{\varepsilon }_{0}}}}\frac{q}{{{{{\left( {AP} \right)}}^{2}}}}$ along $\displaystyle \overrightarrow{{PA}}$

$\displaystyle {{\overrightarrow{E}}_{A}}=\frac{1}{{4\pi {{\varepsilon }_{0}}}}\frac{q}{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}$

$\displaystyle {{\overrightarrow{E}}_{B}}=\frac{1}{{4\pi {{\varepsilon }_{0}}}}\frac{q}{{{{{\left( {BP} \right)}}^{2}}}}$ along $\displaystyle \overrightarrow{{BP}}$

$\displaystyle {{\overrightarrow{E}}_{B}}=\frac{1}{{4\pi {{\varepsilon }_{0}}}}\frac{q}{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}$

The resultant intensity is the vector sum of the intensities along $\displaystyle \overrightarrow{{PA}}$ and $\displaystyle \overrightarrow{{BP}}$ . E_A and E_B can be

resolved into vertical and horizontal components. The vertical components of E_A and E_B cancel

each other as they are equal and oppositely directed. So the horizontal components add up to give

the resultant field.

E = E_A cos θ + E_B cos θ

E = 2E_A cosθ, as E_A = E_B

Substituting, $\displaystyle \cos \theta =\frac{a}{{{{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}^{{\frac{1}{2}}}}}}$ in the above equation

$\displaystyle E=2{{E}_{A}}\cos \theta =\frac{2}{{4\pi {{\varepsilon }_{0}}}}\frac{q}{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}\frac{a}{{{{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}^{{\frac{1}{2}}}}}}$

$\displaystyle E=\frac{{kp}}{{{{{\left( {{{r}^{2}}+{{a}^{2}}} \right)}}^{{\frac{3}{2}}}}}}$

Along $\displaystyle \overrightarrow{{BA}}$ (As p = q × 2a)

As a special case,

If a² << r² then, $\displaystyle E=\frac{{kp}}{{{{r}^{3}}}}$ Along $\displaystyle \overrightarrow{{BA}}$

Electric field intensity at an axial point is twice the electric intensity on the equatorial line. Direction of field will be against the direction of dipole moment.

5. Use Gauss’ law to derive the expression for the electric field due to a straight uniformly charged infinite line of charge density λ C/m.

Sol. To calculate the electric field, imagine a cylindrical Gaussian surface, since the field is everywhere radial, flux through two ends of the cylindrical Gaussian surface is zero.

At cylindrical part of the surface electric field $\displaystyle \overrightarrow{E}$ is normal to the surface at every point and its magnitude is constant.