Electric Charges and Fields

Question 1

Q: Two small identical spheres carry charges of $+5 \, \mu C$ and $-3 \, \mu C$ and are separated by a distance of 10 cm. Calculate the force between the charges and determine its nature.

Solution:
The force between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by Coulomb's law:

$$F = \frac{k \, |q_1 \, q_2|}{r^2}$$

where $k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2$ .

Substitute $q_1 = 5 \times 10^{-6} \, \text{C}$, $q_2 = -3 \times 10^{-6} \, \text{C}$, and $r = 0.1 \, \text{m}$:

$$F = \frac{8.99 \times 10^9 \times |5 \times 10^{-6} \times (-3) \times 10^{-6}|}{(0.1)^2} = \frac{8.99 \times 10^9 \times 15 \times 10^{-12}}{0.01} = 13.485 \, \text{N}$$

The negative sign indicates attraction since one charge is positive and the other is negative. Therefore, the force is attractive with a magnitude of $13.485 \, \text{N}$.

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Question 2

Q: Derive an expression for the electric field at a point on the perpendicular bisector of an electric dipole. Explain why the field decreases with distance faster than for a point charge.

Solution:
Consider an electric dipole with charges $+q$ and $-q$ separated by a distance $2a$. We want the electric field at a point $P$ at a distance $r$ from the center on the perpendicular bisector.

1. The distance of $P$ from each charge is $\sqrt{r^2 + a^2}$ .
2. The field due to $+q$ at $P$ is: $$E_+ = \frac{kq}{(r^2 + a^2)}$$ Similarly, the field due to $-q$ is: $$E_- = \frac{kq}{(r^2 + a^2)}$$

Because the fields are directed oppositely along the line joining $P$ to each charge, the net field at $P$ (perpendicular component) is:

$$E = 2 \cdot E_+ \cdot \frac{a}{\sqrt{r^2 + a^2}} = \frac{2kqa}{(r^2 + a^2)^{3/2}}$$

For large $r$ (far field), $r^2 \gg a^2$ , so:

$$E \approx \frac{2kqa}{r^3}$$

This shows that the field of a dipole decreases as $1/r^3$, faster than the $1/r^2$ dependence for a point charge.

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Question 3

Q: A point charge of $+4 \, \mu C$ is placed at the center of a cube with side length 20 cm. Calculate the electric flux through one face of the cube.

Solution:
By Gauss’s law, the total flux through a closed surface enclosing a charge $Q$ is:

$$\Phi = \frac{Q}{\epsilon_0}$$

Since the cube has six faces and the charge is symmetrically placed at the center, the flux through each face is:

$$\Phi_{\text{face}} = \frac{\Phi}{6} = \frac{Q}{6 \epsilon_0}$$

Substitute $Q = 4 \times 10^{-6} \, \text{C}$ and $\epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2$ :

$$\Phi_{\text{face}} = \frac{4 \times 10^{-6}}{6 \times 8.85 \times 10^{-12}} = 7.53 \times 10^4 \, \text{Nm}^2/\text{C}$$

Thus, the flux through one face of the cube is $7.53 \times 10^4 \, \text{Nm}^2/\text{C}$.

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Question 4

Q: Explain the physical significance of electric flux and its dependence on the orientation of a surface in an electric field. Use an example of a flat sheet placed at different angles to a uniform field.

Solution:
Electric flux, $\Phi = \vec{E} \cdot \vec{A} = E A \cos \theta$ , measures the number of electric field lines passing through a surface. It depends on:

1. Field Strength $E$: Stronger fields have more lines.
2. Surface Area $A$: Larger areas intersect more lines.
3. Angle $\theta$: Maximum flux when perpendicular ($\cos 0^\circ = 1$ ); zero when parallel ($\cos 90^\circ = 0$ ).

Consider a flat sheet in a uniform field:

• At $\theta = 0^\circ$ , flux is maximum as $\Phi = EA$.
• At $\theta = 90^\circ$ , flux is zero because the field lines are parallel to the sheet. This shows flux varies with orientation, reflecting the alignment of the surface with the field lines.

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Question 5

Q: Calculate the potential at a point on the axis of an electric dipole at a distance $r$ from its center. Assume $r \gg a$, where $2a$ is the separation of charges.

Solution:
The potential $V$ at a point on the axis of a dipole (at distance $r$) is given by:

$$V = \frac{k \, p}{r^2}$$

where $p = q \times 2a$ is the dipole moment.

1. The potential due to $+q$ at distance $r - a$ is: $$V_+ = \frac{kq}{r - a} \approx \frac{kq}{r} \left(1 + \frac{a}{r}\right)$$
2. For $-q$ at distance $r + a$ : $$V_- = -\frac{kq}{r + a} \approx -\frac{kq}{r} \left(1 - \frac{a}{r}\right)$$
3. The net potential: $$V = V_+ + V_- = \frac{2kqa}{r^2} = \frac{k \, p}{r^2}$$

Thus, the potential on the axis of a dipole is $V = \frac{k \, p}{r^2}$ , indicating it decreases as $1/r^2$ .

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Question 6

Q: Derive the expression for the torque experienced by a dipole in a uniform electric field and describe its behavior as the angle with the field varies.

Solution:
For a dipole with charges $+q$ and $-q$ separated by distance $2a$ and dipole moment $\vec{p} = q \cdot 2a$ , placed in a uniform electric field $\vec{E}$ , the force on $+q$ and $-q$ is $F = qE$ .

1. Torque $\tau$ about the center:

   $$\tau = pE \sin \theta$$

   where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$ .

2. As $\theta$ changes:

   • At $\theta = 0^\circ$ : $\tau = 0$ (aligned with $E$ ).
   • At $\theta = 90^\circ$ : $\tau = pE$ (maximum).

Thus, torque varies sinusoidally with $\theta$, being maximum when perpendicular to the field.

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Question 7

Q: Two point charges, $q_1 = +2 \, \mu C$ and $q_2 = -3 \, \mu C$ , are placed 15 cm apart in a vacuum. Determine the position along the line joining the two charges where the electric field is zero.

Solution:
The electric field will be zero at a point where the magnitudes of the electric fields due to both charges are equal but in opposite directions. Let this point be at a distance $x$ from $q_1$ on the line joining $q_1$ and $q_2$ .

For electric field to be zero at a point $x$ from $q_1$ , the field due to $q_1$ must equal the field due to $q_2$ in magnitude:

$$\frac{k |q_1|}{x^2} = \frac{k |q_2|}{(0.15 - x)^2}$$

Cancel $k$ and rearrange:

$$\frac{2 \times 10^{-6}}{x^2} = \frac{3 \times 10^{-6}}{(0.15 - x)^2}$$

Taking the square root on both sides:

$$\frac{\sqrt{2}}{x} = \frac{\sqrt{3}}{0.15 - x}$$

Cross-multiply:

$$\sqrt{2} (0.15 - x) = \sqrt{3} x$$

Solving for $x$ , we get the position where the electric field is zero.

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Question 8

Q: Calculate the work done in moving a 2 µC charge from a point at potential $V_A = 50 \, \text{V}$ to a point at $V_B = 100 \, \text{V}$ .

Solution:
The work done $W$ in moving a charge $q$ between two points $A$ and $B$ with potential difference $V_B - V_A$ is given by:

$$W = q (V_B - V_A)$$

Given $q = 2 \times 10^{-6} \, \text{C}$ , $V_A = 50 \, \text{V}$ , and $V_B = 100 \, \text{V}$ :

$$W = 2 \times 10^{-6} \times (100 - 50) = 2 \times 10^{-6} \times 50 = 0.1 \, \text{mJ}$$

So, the work done is 0.1 mJ.

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Question 9

Q: Derive an expression for the energy stored in a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ .

Solution:
The potential energy $U$ of a system of two charges $q_1$ and $q_2$ separated by distance $r$ is given by:

$$U = \frac{k q_1 q_2}{r}$$

where $k = \frac{1}{4 \pi \epsilon_0}$ .

This represents the work required to bring $q_2$ from infinity to a distance $r$ from $q_1$ . Thus, $U = \frac{k q_1 q_2}{r}$ is the expression for the potential energy of two point charges.

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Question 10

Q: A large plane sheet has a uniform charge density $\sigma$ . Derive the expression for the electric field near the surface of the sheet using Gauss's law.

Solution:
Consider a Gaussian surface in the form of a cylinder with one face parallel to the sheet.

1. By symmetry, the electric field $E$ is perpendicular to the sheet and uniform over the Gaussian surface.
2. The total flux through the Gaussian surface is $\Phi = E \times 2A$ (since flux is only through the two faces of the cylinder).
3. By Gauss’s law: $$\Phi = \frac{q_{\text{enc}}}{\epsilon_0}$$ where $q_{\text{enc}} = \sigma A$ .

Thus:

$$E \times 2A = \frac{\sigma A}{\epsilon_0} \Rightarrow E = \frac{\sigma}{2 \epsilon_0}$$

So, the electric field near the surface of a uniformly charged plane sheet is $E = \frac{\sigma}{2 \epsilon_0}$ .

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Question 11

Q: Show that for a uniformly charged spherical shell, the electric field inside the shell is zero.

Solution:
Using Gauss's law, consider a Gaussian surface inside the shell. Since there’s no enclosed charge inside a shell, the net electric flux is zero.

1. According to Gauss's law: $$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0} = 0$$
2. Since $\vec{E}$ is constant over the Gaussian surface and there is no enclosed charge: $$E \cdot A = 0 \Rightarrow E = 0$$

Thus, the electric field inside a uniformly charged spherical shell is zero.

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Question 12

Q: Calculate the electric field at a point on the axis of a ring of radius $R$ carrying a uniform charge $Q$ , at a distance $x$ from its center.

Solution:

1. For a ring with total charge $Q$ , the element of charge $dq$ produces an electric field component along the axis.
2. By symmetry, only the component along the axis contributes to the net field.

The field $dE$ due to $dq$ at distance $x$ from the center of the ring on the axis is:

$$dE = \frac{kdq}{r^2} \cos \theta$$

where $r = \sqrt{R^2 + x^2}$ and $\cos \theta = \frac{x}{\sqrt{R^2 + x^2}}$ .

The total field:

$$E = \frac{kQx}{(R^2 + x^2)^{3/2}}$$

Thus, the electric field at a point on the axis is $E = \frac{kQx}{(R^2 + x^2)^{3/2}}$ .

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Question 13

Q: A uniformly charged insulating sphere of radius $R$ has a total charge $Q$ distributed uniformly throughout its volume. Derive the expression for the electric field inside the sphere at a distance $r$ from the center.

Solution:
Using Gauss’s law, consider a Gaussian surface inside the sphere at distance $r$ .

1. The charge enclosed within radius $r$ is:

   $$q_{\text{enc}} = \rho \frac{4}{3} \pi r^3$$

   where $\rho = \frac{Q}{\frac{4}{3} \pi R^3}$ .

2. Substitute $\rho$:

   $$q_{\text{enc}} = Q \frac{r^3}{R^3}$$

3. Using Gauss’s law:

   $$E \times 4 \pi r^2 = \frac{q_{\text{enc}}}{\epsilon_0}$$

Substitute $q_{\text{enc}}$ :

$$E \times 4 \pi r^2 = \frac{Qr^3}{R^3 \epsilon_0} \Rightarrow E = \frac{Qr}{4 \pi \epsilon_0 R^3}$$

Thus, the electric field inside a uniformly charged sphere is $E = \frac{Qr}{4 \pi \epsilon_0 R^3}$ .

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Question 14

Q: Determine the potential due to a continuous line of charge with linear charge density $\lambda$ at a perpendicular distance $r$ .

Solution:
The electric potential $V$ at a distance $r$ from an infinite line of charge (with uniform linear charge density $\lambda$ ) can be derived by integrating the electric field along the perpendicular distance.

For an infinite line charge, the field is:

$$E = \frac{\lambda}{2 \pi \epsilon_0 r}$$

The potential difference between two points (taking reference at infinity) is:

$$V = - \int_{\infty}^r \frac{\lambda}{2 \pi \epsilon_0 r} \, dr$$

This leads to:

$$V = -\frac{\lambda}{2 \pi \epsilon_0} \ln \frac{r}{r_0}$$

Thus, the potential due to an infinite line of charge is proportional to $\ln \frac{1}{r}$ , which shows a logarithmic dependency.

Question 15

Q: A solid insulating sphere of radius $R$ carries a uniform charge $Q$ . Derive the expression for the potential $V(r)$ at a distance $r$ from the center:

1. For $r > R$ (outside the sphere).
2. For $r \leq R$ (inside the sphere).

Solution:

1. For $r > R$ (outside the sphere):

The sphere behaves like a point charge at its center for points outside it. The potential is given by:

$$V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$$

2. For $r \leq R$ (inside the sphere):

Using the electric field inside the sphere derived previously:

$$E = \frac{Qr}{4\pi\epsilon_0 R^3}$$

The potential at any point $r$ is obtained by integrating the field from $R$ to $r$ :

$$V(r) = V(R) - \int_r^R E \, dr$$

Here, $V(R) = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$ , and substituting $E$ :

$$V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} - \int_r^R \frac{Qr}{4\pi\epsilon_0 R^3} \, dr$$ $$V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} - \frac{Q}{4\pi\epsilon_0 R^3} \left[ \frac{r^2}{2} \right]_r^R$$

After solving:

$$V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} - \frac{Q}{4\pi\epsilon_0 R^3} \left(\frac{R^2}{2} - \frac{r^2}{2}\right)$$ $$V(r) = \frac{1}{4\pi\epsilon_0} \left[\frac{3Q}{2R} - \frac{Qr^2}{2R^3}\right]$$

Thus:

$$\boxed{V(r) = \frac{1}{4\pi\epsilon_0} \left[\frac{3Q}{2R} - \frac{Qr^2}{2R^3}\right]}$$

for $r \leq R$ .

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Question 16

Q: Prove that the electric field near the midpoint of two equal charges but opposite in sign forms a dipole and show its dependence on distance $r$ .

Solution:

Consider a dipole with charges $+q$ and $-q$ separated by a distance $2a$ .

At a point $P$ on the axial line at distance $r$ from the midpoint of the dipole:

Electric field due to the dipole is:

$$E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2p}{r^3}, \quad \text{where } p = q \cdot 2a$$

Here, $p$ is the dipole moment. This shows that the electric field of a dipole decreases as $r^{-3}$ .

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Question 17

Q: Derive the expression for the torque acting on an electric dipole placed in a uniform electric field $\vec{E}$ .

Solution:

An electric dipole in a uniform field experiences forces $+q\vec{E}$ and $-q\vec{E}$ , separated by $2a$ . These forces form a couple causing torque.

Torque is given by:

$$\vec{\tau} = \vec{p} \times \vec{E}$$

where $\vec{p} = q \cdot 2a$ .

Magnitude of torque:

$$\tau = pE \sin \theta$$

Thus, the torque acting on a dipole is $\tau = pE \sin \theta$ , where $\theta$ is the angle between $\vec{p}$ and $\vec{E}$ .

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Question 18

Q: A point charge $Q$ is placed at the center of a hollow spherical shell of radius $R$ and negligible thickness. What is the flux through the shell?

Solution:

By Gauss’s law:

$$\Phi = \frac{q_{\text{enc}}}{\epsilon_0}$$

Here, the charge $Q$ is enclosed, so:

$$\Phi = \frac{Q}{\epsilon_0}$$

Thus, the flux through the shell is $\boxed{\Phi = \frac{Q}{\epsilon_0}}$ .

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Question 19

Q: Derive the expression for the electric field due to a uniformly charged infinite plane sheet using Gauss's law.

Solution:

1. Take a Gaussian surface in the form of a cylinder, with its axis perpendicular to the sheet.
2. By symmetry, the field is uniform and perpendicular to the sheet.
3. Using Gauss's law:

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\epsilon_0}$$

Here, $q_{\text{enc}} = \sigma A$ , and flux through two ends of the cylinder is $E \cdot 2A$ .

$$E \cdot 2A = \frac{\sigma A}{\epsilon_0} \Rightarrow E = \frac{\sigma}{2\epsilon_0}$$

Thus, $\boxed{E = \frac{\sigma}{2\epsilon_0}}$ .

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Question 20

Q: A charge is distributed uniformly along a semicircular arc of radius $R$ . Derive the expression for the electric field at the center of the arc.

Solution:

1. Divide the arc into infinitesimal elements $dq = \lambda dl$ , where $\lambda = \frac{q}{\pi R}$ .
2. The electric field due to each element at the center has components:
   • Radial components cancel out due to symmetry.
   • Only horizontal components add up.

$$E_x = \int dE \cos \theta$$

Substituting $dE = \frac{kdq}{R^2}$ and $dq = \lambda R d\theta$ :

$$E_x = \int_{-\pi/2}^{\pi/2} \frac{k\lambda R}{R^2} \cos \theta d\theta$$ $$E_x = \frac{k\lambda}{R} \int_{-\pi/2}^{\pi/2} \cos \theta d\theta = \frac{k\lambda}{R} [\sin \theta]_{-\pi/2}^{\pi/2}$$ $$E_x = \frac{k\lambda}{R} (1 - (-1)) = \frac{2k\lambda}{R}$$

Substituting $\lambda = \frac{q}{\pi R}$ :

$$E_x = \frac{2kq}{\pi R^2}$$

Thus, the electric field at the center is $\boxed{\frac{2kq}{\pi R^2}}$ .

Question 21

Q: A parallel plate capacitor is charged and then disconnected from the battery. Now the plates are pulled apart. Explain the changes in the capacitance, charge, and potential difference across the plates.

Solution:
When the capacitor is disconnected from the battery, the charge $Q$ on the plates remains constant. The capacitance $C$ depends on the area of the plates and the distance between them. The following effects are observed when the plates are pulled apart:

1. Capacitance:

   $$C = \frac{\epsilon_0 A}{d}$$

   where $A$ is the area of the plates and $d$ is the separation. As the plates are pulled apart, $d$ increases, so the capacitance decreases.

2. Charge:
   Since the capacitor is disconnected from the battery, the charge $Q$ remains constant.

3. Potential Difference:
   The potential difference $V$ is given by:

   $$V = \frac{Q}{C}$$

   As the capacitance decreases, the potential difference increases.

Thus, pulling the plates apart increases the potential difference while the charge remains constant and the capacitance decreases.

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Question 22

Q: A charge $Q$ is placed at the center of a spherical shell of radius $R$ . What is the electric field at points inside and outside the shell?

Solution:
By applying Gauss’s Law:

1. Outside the spherical shell (for $r > R$ ):
   The spherical shell behaves like a point charge at the center. Thus, the electric field at a distance $r$ from the center (outside the shell) is given by:

   $$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$$

2. Inside the spherical shell (for $r < R$ ):
   The charge enclosed by a Gaussian surface inside the shell is zero, so by Gauss’s law:

   $$E = 0$$

Thus, the electric field is $E = 0$ inside the shell and $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$ outside the shell.

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Question 23

Q: A charged particle enters a uniform magnetic field with velocity $\vec{v}$ perpendicular to the field. Derive the expression for the radius of the circular path followed by the particle.

Solution:
For a charged particle moving in a magnetic field, the magnetic force acts as the centripetal force. The magnetic force is given by:

$$F_B = qvB$$

where $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field strength. The centripetal force required for circular motion is:

$$F_c = \frac{mv^2}{r}$$

Equating the magnetic force to the centripetal force:

$$qvB = \frac{mv^2}{r}$$

Solving for $r$ :

$$r = \frac{mv}{qB}$$

Thus, the radius of the circular path is $\boxed{r = \frac{mv}{qB}}$ .

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Question 24

Q: A uniformly charged ring of radius $R$ carries a total charge $Q$ . Derive the expression for the electric field at a point along the axis of the ring at distance $x$ from its center.

Solution:
Consider an element of charge $dq$ on the ring. The distance from $dq$ to the point on the axis is $\sqrt{R^2 + x^2}$ . The electric field due to $dq$ is directed radially outward, but the horizontal components cancel due to symmetry. Only the vertical components contribute to the total field.

The electric field due to $dq$ is:

$$dE = \frac{1}{4\pi\epsilon_0} \frac{dq}{(R^2 + x^2)}$$

The vertical component of $dE$ is:

$$dE_{\text{vertical}} = dE \cdot \frac{x}{\sqrt{R^2 + x^2}}$$

Now, integrate over the entire ring to get the total electric field at distance $x$ along the axis:

$$E_x = \frac{1}{4\pi\epsilon_0} \frac{Q}{(R^2 + x^2)^{3/2}} \cdot x$$

Thus, the electric field at a distance $x$ along the axis of the ring is:

$$\boxed{E_x = \frac{1}{4\pi\epsilon_0} \frac{Qx}{(R^2 + x^2)^{3/2}}}$$

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Question 25

Q: Derive the expression for the potential due to a point charge $Q$ at a distance $r$ from the charge.

Solution:
The electric field due to a point charge $Q$ is given by:

$$E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$$

The potential at a distance $r$ from the charge is the negative integral of the electric field:

$$V(r) = -\int_{\infty}^{r} E \, dr$$

Substituting for $E$ :

$$V(r) = -\int_{\infty}^{r} \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} \, dr$$

Solving the integral:

$$V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$$

Thus, the potential due to a point charge at a distance $r$ is:

$$\boxed{V(r) = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}}$$

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Question 26

Q: Two point charges $+q$ and $-q$ are placed at a separation $2a$ . Calculate the electric field at a point on the perpendicular bisector of the line joining the two charges.

Solution:
At any point on the perpendicular bisector of the line joining the charges, the contributions from both charges will have the same magnitude but opposite directions in the horizontal component, so they cancel out. The vertical components add up.

1. The electric field due to each charge at a distance $r$ from the point on the perpendicular bisector is:

$$E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$$

2. The vertical components of the electric field add up. If $\theta$ is the angle between the line connecting the point and the charge, the total electric field is:

$$E_{\text{total}} = 2E \sin \theta$$

Since $\sin \theta = \frac{a}{\sqrt{a^2 + x^2}}$ , where $x$ is the distance from the center, the total field becomes:

$$E = \frac{1}{4\pi\epsilon_0} \frac{2q a}{(a^2 + x^2)^{3/2}}$$

Thus, the electric field at a point on the perpendicular bisector is:

$$\boxed{E = \frac{1}{4\pi\epsilon_0} \frac{2q a}{(a^2 + x^2)^{3/2}}}$$

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Question 27

Q: What is the relation between the electric field and electric potential in electrostatics? Derive the expression for the electric field from the electric potential in one dimension.

Solution:
The electric field is the negative gradient of the electric potential:

$$E = -\frac{dV}{dx}$$

In one dimension, the electric field is the negative rate of change of potential with respect to position $x$ . If $V(x)$ is the potential as a function of $x$ , the electric field is:

$$E(x) = - \frac{dV}{dx}$$

Thus, the electric field can be derived from the potential by differentiating the potential with respect to distance.

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Question 28

Q: If the electric potential at a point in space is zero, does it mean that the electric field at that point is also zero? Explain with examples.

Solution:
No, a zero potential does not necessarily imply a zero electric field. The electric field is related to the rate of change of potential with distance.

• For example, around a point charge, the potential is zero at infinity, but the electric field is not zero at any finite point near the charge.
• A uniform electric field can also have a constant potential, yet the electric field is non-zero.

Thus, a zero potential does not imply a zero electric field.

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Question 29

Q: Show that the potential due to a uniformly charged infinite plane sheet is constant everywhere and does not depend on the distance from the sheet.

Solution:
Consider an infinite plane sheet with a uniform charge distribution. The electric field due to the plane sheet is constant and does not depend on the distance from the sheet. This can be derived using Gauss's law:

$$E = \frac{\sigma}{2\epsilon_0}$$

where $\sigma$ is the surface charge density.

The potential at a distance $x$ from the sheet is:

$$V(x) = - \int_{0}^{x} E \, dx = - \int_{0}^{x} \frac{\sigma}{2\epsilon_0} \, dx = \frac{\sigma x}{2\epsilon_0}$$

Thus, the potential increases linearly with distance, but the electric field is constant and does not depend on distance. Hence, the potential difference is proportional to the distance from the plane, and the electric field remains uniform.

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Question 30

Q: Derive the expression for the electric potential energy stored in a capacitor of capacitance $C$ and charge $Q$ .

Solution:
The work done to charge a capacitor is the potential energy stored in the capacitor. The voltage across a capacitor is related to the charge by:

$$V = \frac{Q}{C}$$

The infinitesimal work done to move an infinitesimal charge $dq$ onto the capacitor is:

$$dW = V \, dq = \frac{Q}{C} \, dq$$

Integrating this from 0 to $Q$ gives the total work (and potential energy):

$$W = \int_0^Q \frac{Q}{C} \, dq = \frac{Q^2}{2C}$$

Thus, the potential energy stored in the capacitor is:

$$\boxed{U = \frac{Q^2}{2C}}$$