Electrostatics

An oil drop with a radius of $1 \, \mu \mathrm{m}$ is held stationary in a constant electric field of $3.65 \times 10^{4} \, \mathrm{N/C}$ due to excess electrons on it. The oil drop's density is $1.26 \, \mathrm{g/cm}^{3}$. To find the number of excess electrons on the oil drop, consider the following details:

Convert the given measurements:

Radius of the drop, $ r = 1 \, \mu \mathrm{m} = 10^{-6} \mathrm{~m} $.

Density of the oil, $\rho_{\text{oil}}$, is converted to:

$ \rho_{\text{oil}} = 1.26 \, \mathrm{g/cm}^{3} = 1.26 \times 10^{3} \, \mathrm{kg/m}^{3} $

Since the droplet is stationary, the gravitational force acting downward is balanced by the electric force acting upward:

$ \text{Weight of the droplet} = \text{Force due to the electric field} $

Mathematically, this can be expressed as:

$ \frac{4}{3} \pi r^{3} \rho_{\text{oil}} \cdot g = qE $

where $ q $ is the charge due to excess electrons. If $ n $ is the number of excess electrons, then:

$ q = ne $

Substitute this into the previous equation:

$ \frac{4}{3} \pi r^{3} \rho_{\text{oil}} \cdot g = neE $

Solving for $ n $:

$ n = \frac{4 \pi r^{3} \rho_{\text{oil}} g}{3eE} $

Substitute the known values into the equation to calculate $ n $:

$ \begin{aligned} n & = \frac{4 \times 3.14 \times (10^{-6})^{3} \times 1.26 \times 10^{3} \times 10}{3 \times 1.6 \times 10^{-19} \times 3.65 \times 10^{4}} \\ & = 9.03 \approx 9 \end{aligned} $

Thus, there are approximately 9 excess electrons on the oil drop.

When eight small liquid drops combine to form one large drop, the potential of the large drop is given as 20 V. We need to find the potential of each individual small drop.

Solution

Volume Relationship:

The total volume of the eight small drops is equal to the volume of the one large drop. This relationship is expressed as:

$ \left(\frac{4}{3} \pi r^3\right) \times 8 = \frac{4}{3} \pi R^3 $

Simplifying gives:

$ 2r = R \quad \ldots \text{(i)} $

This equation indicates the radius of the large drop $ R $ is twice the radius of one small drop $ r $.

Charge Conservation:

The total charge of the eight small drops is equal to the charge of the large drop:

$ 8q = Q \quad \ldots \text{(ii)} $

Potential Derivation:

The potential for one small drop ($ V' $) and the large drop ($ V $) are given by:

$ V' = \frac{q}{4 \pi \varepsilon_0 r} $

$ V = \frac{Q}{4 \pi \varepsilon_0 R} $

Finding the Ratio:

Using the above expressions, we derive:

$ \frac{V'}{V} = \frac{q}{Q} \times \frac{R}{r} $

Substituting from equations (i) and (ii), we get:

$ \frac{V'}{20} = \frac{q}{8q} \times \frac{2r}{r} $

This simplifies to:

$ \frac{V'}{20} = \frac{1}{4} $

Calculating $ V' $:

Solving for $ V' $, we find:

$ V' = 5 \, \text{V} $

Thus, the potential of each single small liquid drop is 5 V.

To determine how many electrons have been removed from a neutral dust particle suspended in an electric field, let's go through the calculations step by step.

Given:

Mass of the dust particle, $ m = 4 \times 10^{-12} \text{ mg} $

Conversion to kilograms:

$ m = 4 \times 10^{-12} \times 10^{-3} \text{ g} = 4 \times 10^{-12} \times 10^{-3} \times 10^{-3} \text{ kg} = 4 \times 10^{-18} \text{ kg} $

Electric field strength, $ E = 50 \text{ N/C} $

Gravitational acceleration, $ g = 10 \text{ m/s}^2 $

Calculate the Weight:

The weight $ W $ of the dust particle is given by:

$ W = m \times g = 4 \times 10^{-18} \times 10 = 4 \times 10^{-17} \text{ N} $

Equating Electric Force to Weight:

The electric force $ F_e $ experienced by the dust particle is:

$ F_e = q \times E $

Where $ q = n \times e $, with $ e = 1.6 \times 10^{-19} \text{ C} $ as the charge of one electron, and $ n $ being the number of electrons removed.

At equilibrium, the electric force equals the weight of the particle:

$ n \times e \times E = W $

$ n \times 1.6 \times 10^{-19} \times 50 = 4 \times 10^{-17} $

Solving for $ n $:

$ n = \frac{4 \times 10^{-17}}{1.6 \times 10^{-19} \times 50} $

$ n = \frac{400}{80} = 5 $

Therefore, 5 electrons were removed from the neutral dust particle.

The electric field $\mathbf{E}$ at point $(30, 30, 0)$ due to a charge of $0.008 \, \mu \mathrm{C}$ placed at the origin can be calculated using the formula:

$ \mathbf{E} = \frac{K q}{r^3} \cdot \mathbf{r} $

Step 1: Calculate the distance $r$:

$ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} $

Substitute the given coordinates:

$ r = \sqrt{30^2 + 30^2 + 0^2} = 30 \sqrt{2} \, \text{cm} = 30 \sqrt{2} \times 10^{-2} \, \text{m} $

Step 2: Convert charge $q$ to Coulombs:

$ q = 8 \times 10^{-3} \, \mu\mathrm{C} = 8 \times 10^{-3} \times 10^{-6} \, \text{C} $

Step 3: The position vector $\mathbf{r}$ is:

$ \mathbf{r} = (30 \hat{\mathbf{i}} + 30 \hat{\mathbf{j}}) \times 10^{-2} \, \text{m} $

Step 4: Substitute into the electric field formula:

$ \mathbf{E} = \frac{9 \times 10^9 \times 8 \times 10^{-3} \times 10^{-6}}{\left(30 \sqrt{2} \times 10^{-2}\right)^3} \times (30 \hat{\mathbf{i}} + 30 \hat{\mathbf{j}}) \times 10^{-2} $

Calculation:

$ \mathbf{E} = \frac{9 \times 8 \times 10^9 \times 10^{-11}}{27 \times 2 \sqrt{2} \times 10^{-6} \times 10^3} \times 30(\hat{\mathbf{i}} + \hat{\mathbf{j}}) $

Simplify the expression:

$ \mathbf{E} = \frac{9 \times 8 \times 10^2 \times 3}{27 \times 2 \sqrt{2}}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) $

$ \mathbf{E} = 200 \sqrt{2}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \, \text{NC}^{-1} $

Thus, the electric field at the given point is $200 \sqrt{2}(\hat{\mathbf{i}} + \hat{\mathbf{j}}) \, \mathrm{NC}^{-1}$.

Electric potential at centre R = 0

V_centre = 1.5 V_surface

where,

${V_0} = {{kQ} \over R}$

${V_{inside}} = {{kQ} \over {2R}}\left( {3 - {{{r^2}} \over {{R^2}}}} \right)$

${V_{outside}} = {{kQ} \over r} \propto {1 \over r}$

Consider the equilibrium of charge Q at corner A

${F_B} = {F_D} = {1 \over {4\pi {\varepsilon _0}}}{{{Q^2}} \over {{a^2}}} = {{k{Q^2}} \over {{a^2}}}$ [$\because$ ${1 \over {4\pi {\varepsilon _0}}} = k$]

${F_C} = {{k{Q^2}} \over {{{(\sqrt 2 a)}^2}}}$ [$\because$ $AC = \sqrt 2 a$]

$ = {{k{Q^2}} \over {2{a^2}}}$

${F_O} = {{k\,Qq} \over {{{\left( {{a \over {\sqrt 2 }}} \right)}^2}}} = {{k\,2Qq} \over {{a^2}}}$ [$\because$ $OC = {a \over {\sqrt 2 }}$]

For equilibrium of charge Q at corner A, the net force on this charge along AC must be zero.

$\therefore$ ${F_O} = {F_C} + {F_B}\cos 45^\circ + {F_D}\cos 45^\circ $

$ = {F_C} + 2{F_B}\cos 45^\circ $ [$\because$ ${F_B} = {F_D}$]

$ = {F_C} + 2{F_B} \times {1 \over {\sqrt 2 }}$ [$\because$ $\cos 45^\circ = {1 \over {\sqrt 2 }}$]

$ \Rightarrow {{k\,2Qq} \over {{a^2}}} = {{k\,{Q^2}} \over {2{a^2}}} + {{2 \times k{Q^2}} \over {{a^2}}} \times {1 \over {\sqrt 2 }}$

$ \Rightarrow 2Qq = {{{Q^2}} \over 2} + {{{Q^2}} \over {\sqrt 2 }} \times 2$

$ \Rightarrow 2q = {Q \over 2} + Q\sqrt 2 $

$ \Rightarrow q = {Q \over 4}(1 + 2\sqrt 2 )$

When a dipole is placed in a non-uniform electric field, opposite charges of the dipole experiences a force due to E. It is as shown in the figure below

These forces will act in opposite direction. Now, since the electric field is not uniform, so the force experienced by the charges will be unequal.

Hence, their will be a net force acting on the dipole, as they do not cancel each other.

Also, forces on the charges are not linear as shown above. So, they will experience a not non-zero torque. Hence, the dipole will experience both translational force and torque in a non-uniform electric field.

Potential at point A,

${V_A} = {1 \over {4\pi {\varepsilon _0}}}.{q \over a}$

Potential at point B,

${V_B} = {1 \over {4\pi {\varepsilon _0}}}.{q \over b}$

Work done in taking a charge Q from A to B,

$W = Q({V_B} - {V_A}) = {{Qq} \over {4\pi {\varepsilon _0}}}\left[ {{1 \over b} - {1 \over a}} \right] = {{Qq} \over {4\pi {\varepsilon _0}}}\left[ {{{a - b} \over {ab}}} \right]$

Given, two small conducting spheres of equal radius (R).

Charge on first sphere, q₁ = +20 $\mu$C

Charge on second sphere, q₂ = $-$40 $\mu$C

$\therefore$ ${F_1} = {{k( + 20)( - 40)} \over {{R^2}}} = {{ - k(800)} \over {{R^2}}}$ ..... (i)

The spheres have equal radii, so their capacities will be same. Now, if they are brought in contact, then the net charge,

q_net = (+20 $-$40) $\mu$C = $-$20 $\mu$C

So, each sphere will have charge, $q = {{{q_{net}}} \over 2} = - 10$ $\mu$C

Now, ${F_2} = {{k(q)(q)} \over {{R^2}}} = {{k( - 10)( - 10)} \over {{R^2}}} = {{k(100)} \over {{R^2}}}$ ..... (ii)

Ratio, ${{{F_1}} \over {{F_2}}} = - {{800} \over {100}} = - 8:1$

Thus, ratio of F₁ to F₂ = $-$8 : 1