Electric Charges and Fields

Given:
- $q_1 = 2 \times 10^{-7}$ C
- $q_2 = 3 \times 10^{-7}$ C
- $r = 30$ cm $= 0.30$ m
- $k = 9 \times 10^9$ N m² C⁻²

Formula (Coulomb's Law):
$$F = k\frac{q_1 q_2}{r^2}$$

Calculation:
$$F = 9 \times 10^9 \times \frac{(2 \times 10^{-7})(3 \times 10^{-7})}{(0.30)^2}$$
$$F = 9 \times 10^9 \times \frac{6 \times 10^{-14}}{9 \times 10^{-2}}$$
$$F = 9 \times 10^9 \times \frac{6}{9} \times 10^{-12}$$
$$F = 6 \times 10^{-3} \text{ N}$$

Result: The force between the two spheres is $6 \times 10^{-3}$ N. Since both charges are positive, the force is repulsive.

Given:
- $q_1 = 0.4\,\mu\text{C} = 0.4 \times 10^{-6}$ C
- $q_2 = -0.8\,\mu\text{C} = -0.8 \times 10^{-6}$ C
- $F = 0.2$ N
- $k = 9 \times 10^9$ N m² C⁻²

(a) Distance between the spheres:

Using Coulomb's law:
$$F = k\frac{|q_1||q_2|}{r^2}$$
$$r^2 = k\frac{|q_1||q_2|}{F}$$
$$r^2 = \frac{9 \times 10^9 \times 0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{0.2}$$
$$r^2 = \frac{9 \times 10^9 \times 3.2 \times 10^{-13}}{0.2}$$
$$r^2 = \frac{2.88 \times 10^{-3}}{0.2} = 1.44 \times 10^{-2}$$
$$r = 0.12 \text{ m} = 12 \text{ cm}$$

(b) Force on the second sphere due to the first:

By Newton's Third Law, the force on the second sphere due to the first is equal in magnitude and opposite in direction to the force on the first sphere due to the second.

$$\boxed{F = 0.2 \text{ N (attractive, directed towards the first sphere)}}$$

Checking dimensions:

- $k = \frac{1}{4\pi\varepsilon_0}$ has dimensions $[\text{M L}^3 \text{T}^{-4} \text{A}^{-2}]$
- $e$ (charge) has dimensions $[\text{A T}]$
- $G$ (gravitational constant) has dimensions $[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$
- $m_e$, $m_p$ (masses) have dimensions $[\text{M}]$

Dimensions of numerator $ke^2$:
$$[\text{M L}^3 \text{T}^{-4} \text{A}^{-2}][\text{A}^2\text{T}^2] = [\text{M L}^3 \text{T}^{-2}]$$

Dimensions of denominator $Gm_em_p$:
$$[\text{M}^{-1}\text{L}^3\text{T}^{-2}][\text{M}][\text{M}] = [\text{M L}^3 \text{T}^{-2}]$$

Since numerator and denominator have the same dimensions, the ratio is dimensionless. ✓

Numerical value:

Using standard values:
- $k = 9 \times 10^9$ N m² C⁻²
- $e = 1.6 \times 10^{-19}$ C
- $G = 6.67 \times 10^{-11}$ N m² kg⁻²
- $m_e = 9.11 \times 10^{-31}$ kg
- $m_p = 1.67 \times 10^{-27}$ kg

$$\frac{ke^2}{Gm_em_p} = \frac{9\times10^9 \times (1.6\times10^{-19})^2}{6.67\times10^{-11}\times9.11\times10^{-31}\times1.67\times10^{-27}}$$

$$= \frac{9\times10^9 \times 2.56\times10^{-38}}{6.67\times9.11\times1.67\times10^{-69}}$$

$$= \frac{2.304\times10^{-28}}{1.016\times10^{-67}} \approx 2.27 \times 10^{39}$$

Significance: This ratio ($\approx 2.4 \times 10^{39}$) represents the ratio of the electrostatic force to the gravitational force between a proton and an electron. It shows that the electric force between an electron and a proton is about $2.4 \times 10^{39}$ times stronger than the gravitational force between them.

(a) Quantisation of electric charge:

The statement means that the electric charge on any body is always an integral multiple of the basic unit of charge $e$ (the charge of an electron or proton, $e = 1.6 \times 10^{-19}$ C).

Mathematically: $q = ne$, where $n = 0, \pm1, \pm2, \pm3, \ldots$

This means charge cannot take arbitrary values — it can only exist in discrete packets of $e$. Charge can neither be created nor destroyed, and it always appears as an integer multiple of $e$.

(b) Why quantisation can be ignored at macroscopic scale:

At the macroscopic level, the charges involved are enormously large compared to the elementary charge $e = 1.6 \times 10^{-19}$ C. For example, even $1\,\mu\text{C} = 10^{-6}$ C corresponds to $n = \frac{10^{-6}}{1.6\times10^{-19}} \approx 6.25 \times 10^{12}$ elementary charges.

When $n$ is such a large number, the discreteness (step size $e$) becomes negligibly small compared to the total charge, and the charge appears to vary continuously. Hence, quantisation can be ignored for macroscopic charges.

Explanation:

Before rubbing, both the glass rod and the silk cloth are electrically neutral — the total charge of the system is zero.

When the glass rod is rubbed with silk, electrons are transferred from the glass rod to the silk cloth. As a result:
- The glass rod loses electrons and acquires a positive charge $+q$.
- The silk cloth gains electrons and acquires a negative charge $-q$.

The charges on the two bodies are equal in magnitude and opposite in sign.

Total charge after rubbing $= +q + (-q) = 0$

This equals the total charge before rubbing (zero). Hence, no charge is created or destroyed — charge is only transferred from one body to another.

This is perfectly consistent with the law of conservation of charge, which states that the total charge of an isolated system remains constant.

Given:
- $q_A = 2\,\mu\text{C}$, $q_B = -5\,\mu\text{C}$, $q_C = 2\,\mu\text{C}$, $q_D = -5\,\mu\text{C}$
- Side of square $= 10$ cm $= 0.1$ m
- Test charge $q_0 = 1\,\mu\text{C}$ at centre O

Distance from each corner to centre:
$$r = \frac{\text{diagonal}}{2} = \frac{0.1\sqrt{2}}{2} = \frac{0.1}{\sqrt{2}} \approx 0.0707 \text{ m}$$

Analysis by symmetry:

The charges at the corners are:
- $q_A = q_C = +2\,\mu\text{C}$ (at diagonally opposite corners A and C)
- $q_B = q_D = -5\,\mu\text{C}$ (at diagonally opposite corners B and D)

Since $q_A = q_C$ and they are at diagonally opposite corners, the forces on $q_0$ due to $q_A$ and $q_C$ are equal in magnitude but opposite in direction (along diagonal AC). They cancel each other.

Similarly, since $q_B = q_D$ and they are at diagonally opposite corners, the forces on $q_0$ due to $q_B$ and $q_D$ are equal in magnitude but opposite in direction (along diagonal BD). They cancel each other.

Therefore, the net force on the charge at the centre is zero.

$$\boxed{F_{\text{net}} = 0}$$

(a) Why field lines cannot have sudden breaks:

An electric field line represents the path along which a positive test charge would move under the influence of the electric field. The electric field exists at every point in space (except at the location of a source charge). If a field line had a sudden break at some point, it would imply that the electric field is zero or undefined at that point (with no charge present there), which is physically impossible in a region free of charges. Therefore, field lines must be continuous curves — they start from positive charges and end on negative charges (or go to infinity), without any breaks in between.

(b) Why two field lines never cross each other:

At any given point in space, the electric field has a unique direction (a single resultant direction). An electric field line is drawn such that the tangent at any point gives the direction of the electric field at that point.

If two field lines were to cross at a point, it would mean that the electric field at that point has two different directions simultaneously — one along the tangent to each field line. This is physically impossible since the electric field at any point can have only one unique direction.

Hence, two field lines can never cross each other.

Given:
- $q_A = +3\,\mu\text{C} = 3 \times 10^{-6}$ C
- $q_B = -3\,\mu\text{C} = -3 \times 10^{-6}$ C
- Distance AB $= 20$ cm, so $r = 10$ cm $= 0.1$ m (distance from each charge to midpoint O)

(a) Electric field at midpoint O:

Electric field at O due to $q_A$ (positive charge, directed away from $q_A$, i.e., from A towards B):
$$E_A = k\frac{q_A}{r^2} = \frac{9\times10^9 \times 3\times10^{-6}}{(0.1)^2} = \frac{2.7\times10^4}{0.01} = 2.7\times10^6 \text{ N/C}$$
Direction: from A to B (let's call this the positive x-direction).

Electric field at O due to $q_B$ (negative charge, directed towards $q_B$, i.e., from O towards B, same direction as $E_A$):
$$E_B = k\frac{|q_B|}{r^2} = \frac{9\times10^9 \times 3\times10^{-6}}{(0.1)^2} = 2.7\times10^6 \text{ N/C}$$
Direction: from A to B (towards $q_B$).

Both fields point in the same direction (from A to B):
$$E_{\text{net}} = E_A + E_B = 2.7\times10^6 + 2.7\times10^6 = 5.4\times10^6 \text{ N/C}$$

The electric field at O is $5.4 \times 10^6$ N/C directed from A to B (from positive to negative charge).

(b) Force on the test charge:

- Test charge $q_0 = -1.5 \times 10^{-9}$ C (negative)

$$F = |q_0| \times E_{\text{net}} = 1.5\times10^{-9} \times 5.4\times10^6 = 8.1\times10^{-3} \text{ N}$$

Since the test charge is negative, the force on it is opposite to the direction of the electric field, i.e., directed from B to A.

The force on the test charge is $8.1 \times 10^{-3}$ N directed from B to A (opposite to the electric field direction).

Given:
- $q_A = +2.5 \times 10^{-7}$ C at A: $(0, 0, -15\text{ cm}) = (0, 0, -0.15\text{ m})$
- $q_B = -2.5 \times 10^{-7}$ C at B: $(0, 0, +15\text{ cm}) = (0, 0, +0.15\text{ m})$

Total charge:
$$q_{\text{total}} = q_A + q_B = 2.5\times10^{-7} + (-2.5\times10^{-7}) = 0$$

Electric dipole moment:

The dipole moment vector $\mathbf{p}$ is directed from the negative charge to the positive charge.

- Negative charge $q_B$ is at B: $(0, 0, +0.15)$ m
- Positive charge $q_A$ is at A: $(0, 0, -0.15)$ m

So $\mathbf{p}$ points from B to A, i.e., in the negative z-direction.

Magnitude of dipole moment:
$$|\mathbf{p}| = q \times 2a$$
where $q = 2.5 \times 10^{-7}$ C and $2a$ = distance between charges $= 30$ cm $= 0.30$ m.

$$|\mathbf{p}| = 2.5\times10^{-7} \times 0.30 = 7.5\times10^{-8} \text{ C m}$$

The electric dipole moment is $\mathbf{p} = 7.5 \times 10^{-8}$ C m directed along the negative z-axis (from B towards A, i.e., from $-q$ to $+q$).

Given:
- Dipole moment $p = 4 \times 10^{-9}$ C m
- Electric field $E = 5 \times 10^4$ N/C
- Angle $\theta = 30°$

Formula:
$$\tau = pE\sin\theta$$

Calculation:
$$\tau = 4\times10^{-9} \times 5\times10^4 \times \sin 30°$$
$$\tau = 4\times10^{-9} \times 5\times10^4 \times 0.5$$
$$\tau = 4\times10^{-9} \times 2.5\times10^4$$
$$\tau = 10^{-4} \text{ N m}$$

The magnitude of the torque acting on the dipole is $10^{-4}$ N m.

Given:
- Charge on polythene $= -3 \times 10^{-7}$ C
- Charge on electron $e = 1.6 \times 10^{-19}$ C

(a) Number of electrons transferred:

Since polythene has acquired a negative charge, electrons have been transferred from wool to polythene.

$$n = \frac{q}{e} = \frac{3\times10^{-7}}{1.6\times10^{-19}}$$
$$n = 1.875 \times 10^{12} \approx 1.9 \times 10^{12} \text{ electrons}$$

Approximately $1.9 \times 10^{12}$ electrons are transferred from wool to polythene.

(b) Transfer of mass:

Yes, since electrons (which have mass) are transferred from wool to polythene, there is indeed a transfer of mass.

Mass of one electron $m_e = 9.11 \times 10^{-31}$ kg

Mass transferred $= n \times m_e = 1.875\times10^{12} \times 9.11\times10^{-31}$
$$= 1.71 \times 10^{-18} \text{ kg}$$

This is an extremely small (negligible) mass, but technically yes, there is a transfer of mass from wool to polythene.

Given:
- $q_A = q_B = 6.5 \times 10^{-7}$ C
- $r = 50$ cm $= 0.5$ m
- $k = 9 \times 10^9$ N m² C⁻²

(a) Initial force:
$$F = k\frac{q_A q_B}{r^2} = \frac{9\times10^9 \times (6.5\times10^{-7})^2}{(0.5)^2}$$
$$= \frac{9\times10^9 \times 4.225\times10^{-13}}{0.25}$$
$$= \frac{3.8025\times10^{-3}}{0.25}$$
$$= 1.521 \times 10^{-2} \text{ N} \approx 1.52 \times 10^{-2} \text{ N}$$

(b) New force when charge is doubled and distance is halved:

- New charge: $q' = 2 \times 6.5\times10^{-7} = 1.3\times10^{-6}$ C
- New distance: $r' = 0.5/2 = 0.25$ m

$$F' = k\frac{q'^2}{r'^2} = \frac{9\times10^9 \times (1.3\times10^{-6})^2}{(0.25)^2}$$
$$= \frac{9\times10^9 \times 1.69\times10^{-12}}{0.0625}$$
$$= \frac{1.521\times10^{-2}}{0.0625} = 0.2434 \text{ N} \approx 0.243 \text{ N}$$

Alternatively, note that $F \propto \frac{q^2}{r^2}$. When $q \to 2q$ and $r \to r/2$:
$$F' = F \times \frac{(2)^2}{(1/2)^2} = F \times \frac{4}{1/4} = 16F = 16 \times 1.521\times10^{-2} \approx 0.243 \text{ N}$$

The new force of repulsion is approximately $0.243$ N.

Note: The figure shows three particle tracks in a uniform electric field. Based on the standard version of this figure in NCERT:
- Particles 1 and 2 curve towards the positive plate (upward, assuming field points upward)
- Particle 3 curves towards the negative plate

Signs of charges:

- Particle 1: Curves towards the positive plate → it is a negative charge.
- Particle 2: Curves towards the positive plate → it is a negative charge.
- Particle 3: Curves towards the negative plate → it is a positive charge.

*(In the standard NCERT figure: particles 1 and 2 deflect in one direction and particle 3 deflects in the opposite direction.)*

Highest charge-to-mass ratio:

The deflection of a charged particle in an electric field is given by:
$$y = \frac{1}{2}\frac{qE}{m}t^2$$

Greater deflection (more curved track) for the same initial velocity implies a higher $q/m$ ratio.

From the figure, particle 3 shows the greatest curvature (sharpest deflection), so it has the highest charge-to-mass ratio.

Given:
- $\mathbf{E} = 3\times10^3\,\hat{i}$ N/C
- Side of square $= 10$ cm $= 0.1$ m
- Area $A = (0.1)^2 = 0.01$ m²

(a) Plane parallel to the $yz$-plane:

If the plane is parallel to the $yz$-plane, the normal to the plane is along the $x$-axis (i.e., $\hat{n} = \hat{i}$).

Angle between $\mathbf{E}$ and $\hat{n}$: $\theta = 0°$

$$\phi = E \cdot A \cos\theta = 3\times10^3 \times 0.01 \times \cos 0°$$
$$\phi = 3\times10^3 \times 0.01 \times 1 = 30 \text{ N m}^2/\text{C}$$

(b) Normal makes $60°$ with the $x$-axis:

$$\phi = E \cdot A \cos\theta = 3\times10^3 \times 0.01 \times \cos 60°$$
$$\phi = 30 \times 0.5 = 15 \text{ N m}^2/\text{C}$$

Given:
- $\mathbf{E} = 3\times10^3\,\hat{i}$ N/C (uniform field along $x$-axis)
- Cube of side 20 cm with faces parallel to coordinate planes

Concept: For a uniform electric field, the net flux through any closed surface is zero, because the field is the same everywhere and the total charge enclosed is zero.

Detailed reasoning:

For a cube with faces parallel to coordinate planes:
- The two faces perpendicular to the $x$-axis (left and right faces) have normals along $\pm\hat{i}$.
- Flux through right face: $\phi_1 = E \times A = 3\times10^3 \times (0.2)^2 = 3\times10^3 \times 0.04 = 120$ N m²/C (outward)
- Flux through left face: $\phi_2 = -E \times A = -120$ N m²/C (inward, so negative)
- The four faces parallel to the $x$-axis have normals perpendicular to $\mathbf{E}$, so flux through each = 0.

Net flux:
$$\phi_{\text{net}} = \phi_1 + \phi_2 = 120 + (-120) = 0$$

The net flux through the cube is zero. This is consistent with Gauss's law since there is no charge enclosed inside the cube.

Given:
- Net outward flux $\phi = 8.0 \times 10^3$ N m²/C
- $\varepsilon_0 = 8.854 \times 10^{-12}$ C² N⁻¹ m⁻²

(a) Net charge inside the box:

By Gauss's law:
$$\phi = \frac{q_{\text{enc}}}{\varepsilon_0}$$
$$q_{\text{enc}} = \phi \times \varepsilon_0 = 8.0\times10^3 \times 8.854\times10^{-12}$$
$$q_{\text{enc}} = 7.08\times10^{-8} \text{ C} \approx 0.07\,\mu\text{C}$$

The net charge inside the box is approximately $0.07\,\mu$C (or $7.08 \times 10^{-8}$ C).

(b) If net flux were zero:

No, we cannot conclude that there are no charges inside the box.

A zero net flux only means that the net charge enclosed is zero. It is possible that there are equal amounts of positive and negative charges inside the box that cancel each other out, giving a net charge of zero and hence zero net flux. For example, a positive charge $+q$ and a negative charge $-q$ inside the box would give zero net flux, yet charges are present inside.

Given:
- Point charge $q = +10\,\mu\text{C} = 10\times10^{-6}$ C
- The charge is 5 cm above the centre of a square of side 10 cm

Using the hint:

Imagine the square as one face of a cube with edge 10 cm. The point charge is at the centre of this cube (since it is 5 cm above the centre of the bottom face, and the cube has edge 10 cm, so the charge is at the centre).

By Gauss's law, the total flux through the entire cube:
$$\phi_{\text{total}} = \frac{q}{\varepsilon_0} = \frac{10\times10^{-6}}{8.854\times10^{-12}} = 1.13\times10^6 \text{ N m}^2/\text{C}$$

By symmetry, the flux is distributed equally through all 6 faces of the cube.

Flux through one face (the given square):
$$\phi_{\text{square}} = \frac{\phi_{\text{total}}}{6} = \frac{1.13\times10^6}{6}$$
$$\phi_{\text{square}} = \frac{10\times10^{-6}}{6\times8.854\times10^{-12}} \approx 1.88\times10^5 \text{ N m}^2/\text{C}$$

The magnitude of the electric flux through the square is approximately $1.88 \times 10^5$ N m²/C.

Given:
- Point charge $q = 2.0\,\mu\text{C} = 2.0\times10^{-6}$ C at the centre of a cube
- $\varepsilon_0 = 8.854\times10^{-12}$ C² N⁻¹ m⁻²

By Gauss's Law:

The net electric flux through any closed surface depends only on the total charge enclosed, regardless of the shape or size of the surface.

$$\phi = \frac{q_{\text{enc}}}{\varepsilon_0} = \frac{2.0\times10^{-6}}{8.854\times10^{-12}}$$
$$\phi = 2.26\times10^5 \text{ N m}^2/\text{C}$$

The net electric flux through the cubic surface is $2.26 \times 10^5$ N m²/C.

Note: The size of the cube (9.0 cm edge) does not affect the answer — only the enclosed charge matters.

Given:
- Electric flux $\phi = -1.0\times10^3$ N m²/C
- Radius of Gaussian surface $r = 10.0$ cm

(a) Flux when radius is doubled:

By Gauss's law, the flux through a closed surface depends only on the charge enclosed, not on the size or shape of the surface. Doubling the radius does not change the enclosed charge.

$$\phi' = \phi = -1.0\times10^3 \text{ N m}^2/\text{C}$$

The flux remains $-1.0 \times 10^3$ N m²/C.

(b) Value of the point charge:

Using Gauss's law:
$$\phi = \frac{q}{\varepsilon_0}$$
$$q = \phi \times \varepsilon_0 = -1.0\times10^3 \times 8.854\times10^{-12}$$
$$q = -8.854\times10^{-9} \text{ C} \approx -8.85 \text{ nC}$$

The value of the point charge is approximately $-8.85$ nC. The negative sign indicates the charge is negative, which is consistent with the inward (negative) flux.

Given:
- Radius of sphere $R = 10$ cm $= 0.1$ m
- Distance from centre $r = 20$ cm $= 0.2$ m
- Electric field $E = 1.5\times10^3$ N/C (radially inward)
- $\varepsilon_0 = 8.854\times10^{-12}$ C² N⁻¹ m⁻²

Since r > R, the sphere behaves as a point charge at the centre.

Using Gauss's law (or Coulomb's law for a sphere):
$$E = \frac{|q|}{4\pi\varepsilon_0 r^2}$$
$$|q| = E \times 4\pi\varepsilon_0 r^2 = \frac{E \times r^2}{k}$$
$$|q| = \frac{1.5\times10^3 \times (0.2)^2}{9\times10^9}$$
$$|q| = \frac{1.5\times10^3 \times 0.04}{9\times10^9} = \frac{60}{9\times10^9} = 6.67\times10^{-9} \text{ C}$$

Since the electric field points radially inward, the charge on the sphere is negative.

$$q = -6.67\times10^{-9} \text{ C} \approx -6.67 \text{ nC}$$

The net charge on the sphere is approximately $-6.67$ nC.

Given:
- Diameter $= 2.4$ m, so radius $R = 1.2$ m
- Surface charge density $\sigma = 80.0\,\mu\text{C/m}^2 = 80.0\times10^{-6}$ C/m²

(a) Total charge on the sphere:

$$q = \sigma \times 4\pi R^2$$
$$q = 80.0\times10^{-6} \times 4\pi \times (1.2)^2$$
$$q = 80.0\times10^{-6} \times 4\pi \times 1.44$$
$$q = 80.0\times10^{-6} \times 18.096$$
$$q = 1.448\times10^{-3} \text{ C} \approx 1.45\times10^{-3} \text{ C}$$

(b) Total electric flux leaving the surface:

By Gauss's law:
$$\phi = \frac{q}{\varepsilon_0} = \frac{1.448\times10^{-3}}{8.854\times10^{-12}}$$
$$\phi = 1.63\times10^{8} \text{ N m}^2/\text{C}$$

The total charge on the sphere is $\approx 1.45 \times 10^{-3}$ C and the total electric flux leaving the surface is $\approx 1.63 \times 10^8$ N m²/C.

Given:
- Electric field $E = 9\times10^4$ N/C
- Distance $r = 2$ cm $= 0.02$ m
- $\varepsilon_0 = 8.854\times10^{-12}$ C² N⁻¹ m⁻²

Formula for electric field due to an infinite line charge:
$$E = \frac{\lambda}{2\pi\varepsilon_0 r}$$

Solving for linear charge density $\lambda$:
$$\lambda = E \times 2\pi\varepsilon_0 r$$
$$\lambda = 9\times10^4 \times 2\pi \times 8.854\times10^{-12} \times 0.02$$
$$\lambda = 9\times10^4 \times 2 \times 3.1416 \times 8.854\times10^{-12} \times 0.02$$
$$\lambda = 9\times10^4 \times 1.113\times10^{-12}$$

Let me compute step by step:
$$2\pi\varepsilon_0 = 2 \times 3.1416 \times 8.854\times10^{-12} = 5.563\times10^{-11}$$
$$\lambda = 9\times10^4 \times 5.563\times10^{-11} \times 0.02$$
$$\lambda = 9\times10^4 \times 1.113\times10^{-12}$$
$$\lambda = 10^{-7} \text{ C/m} = 0.1\,\mu\text{C/m}$$

The linear charge density is $\lambda = 10^{-7}$ C/m $= 0.1\,\mu$C/m.

Given:
- Surface charge density on each plate: $\sigma = 17.0\times10^{-22}$ C/m²
- One plate has $+\sigma$ and the other has $-\sigma$ on their inner faces.
- $\varepsilon_0 = 8.854\times10^{-12}$ C² N⁻¹ m⁻²

Concept: For two parallel plates with equal and opposite surface charge densities:
- The electric field between the plates adds up.
- The electric field outside the plates cancels out.

Each plate alone produces a field $E = \frac{\sigma}{2\varepsilon_0}$ on each side.

(a) Outer region of the first plate:

In the outer region of the positive plate, the field due to the positive plate points outward ($+$) and the field due to the negative plate points inward (towards the negative plate, which is also outward from the positive plate's perspective — but they cancel).

For two plates with $+\sigma$ and $-\sigma$:
- Outside both plates: fields from the two plates are equal and opposite → they cancel.

$$E_{\text{outer}} = 0$$

(b) Outer region of the second plate:

By the same argument:
$$E_{\text{outer}} = 0$$

(c) Between the plates:

Between the plates, the fields from both plates point in the same direction (from $+$ plate to $-$ plate) and add up:
$$E_{\text{between}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$$
$$E_{\text{between}} = \frac{17.0\times10^{-22}}{8.854\times10^{-12}}$$
$$E_{\text{between}} = 1.92\times10^{-10} \text{ N/C} \approx 1.9\times10^{-10} \text{ N/C}$$

Summary:
- (a) $E = 0$ in the outer region of the first plate
- (b) $E = 0$ in the outer region of the second plate
- (c) $E \approx 1.9 \times 10^{-10}$ N/C between the plates, directed from the positive to the negative plate.