JEE Mains:

1. Calculate the electric field at a point due to a dipole:

• Method: Use the formula:

$$ \vec{E} = \frac{1}{4\pi \epsilon_0} \frac{2p}{r^3} \left[ \hat{r} - 3(\hat{r}\cdot\hat{p}) \hat{p} \right] $$

• (\epsilon_0) is the permittivity of free space,
• (p) is the dipole moment,
• (\hat{r}) is a unit vector pointing from the dipole to the observation point,
• ( \hat{p}) is a unit vector in the direction of the dipole moment.

2. Find the potential energy of a dipole in an electric field:

• Method: Use the formula:

$$ U = - \vec{p}\cdot\vec{E} $$

• (\vec{p}) is the dipole moment,
• (\vec{E}) is the electric field.

3. Determine the torque acting on a dipole in an electric field:

• Method: Use the formula:

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

• (\vec{p}) is the dipole moment,
• (\vec{E}) is the electric field.

4. Calculate the electric field due to a continuous charge distribution:

• Method: Use the formula:

$$ \vec{E} = \int \frac{1}{4\pi \epsilon_0} \frac{\rho(\vec{r}’)}{|\vec{r}-\vec{r}’|} d\vec{r}’ $$

• (\epsilon_0) is the permittivity of free space, -(\rho(\vec{r}’)) is the charge density,
• (\vec{r}) is the position vector of the observation point, -(\vec{r}’) is the position vector of the source point.

5. Find the potential energy of a continuous charge distribution:

• Method: Use the formula:

$$ U = \frac{1}{4\pi \epsilon_0} \iiint \frac{\rho(\vec{r}’)\rho(\vec{r})}{|\vec{r}-\vec{r}’|} d\vec{r}d\vec{r}’$$

• (\epsilon_0) is the permittivity of free space,
• ( \rho(\vec{r}’)\rho(\vec{r})) is the charge density at positions (\vec{r}’) and (\vec{r}), respectively,
• (|\vec{r}-\vec{r}’|) is the distance between positions (\vec{r}’) and (\vec{r}).

6. Determine the electric flux through a surface due to a continuous charge distribution:

• Method: Use the formula:

$$ \Phi = \oint \vec{E} \cdot \hat{n} dA $$

• (\vec{E}) is the electric field,
• (\hat{n}) is a unit vector perpendicular to the surface,
• (dA) is the differential area of the surface.

CBSE Board Exams:

1. Define electric dipole and its dipole moment.

• Electric Dipole: An electric dipole is a pair of equal and opposite charges separated by a small distance.

• Dipole Moment: The dipole moment (\vec{p}) is defined as the product of the magnitude of one of the charges (q) and the separation distance (2a). $$ \vec{p} = q (2\vec{a}) = 2q\vec{a}$$ where (\vec{a}) is a vector pointing from the negative charge to the positive charge.

2. State the expression for the electric field due to a dipole at a point on its axial line and equatorial line.

• Axial Line: The electric field (\vec{E}) due to a dipole at a point on its axial line (which passes through the dipole’s center and is perpendicular to the dipole moment) is given by: $$ \vec{E} = \frac{1}{4\pi \epsilon_0} \frac{2p}{r^3}$$

• Equatorial Line: The electric field (\vec{E}) due to a dipole at a point on its equatorial line (which is perpendicular to the dipole’s axis and passes through the dipole’s center) is given by: $$ \vec{E} = \frac{1}{4\pi \epsilon_0} \frac{p}{r^3} $$ where (\epsilon_0) is the permittivity of free space, (r) is the distance from the dipole, and (p) is the magnitude of the dipole moment.

3. Explain the concept of continuous charge distribution and its charge density.

• A continuous charge distribution can be treated as several discrete point charges close to each other. In such a distribution the magnitude of the charge changes smoothly throughout the region in which the distribution is said to be continuous.

• Charge Density: The charge density (\rho) at a point in a continuous charge distribution is defined as the amount of charge (dq) in a small volume (dV) divided by the volume. Thus, $$\rho = \frac{dq}{dV} $$ where (\rho) is measured in C/m³.

4. Derive the expression for the electric field due to a continuous charge distribution.

Considering the charge distribution as made up of small point charges, we first write the electric field due to a tiny charge (dq) as, $$ d\vec{E}=\frac{1}{4\pi \epsilon_0}\frac{dq}{r^2}\hat{r} $$ where (\hat{r}) is a unit vector pointing from the charge (dq) to the observation point (P) at which the field is being calculated. Adding the contributions from all such point charges, we obtain the field at (P) as, $$\vec{E}=\frac{1}{4\pi \epsilon_0}\int \frac{\rho(\overrightarrow{r}’)}{|\overrightarrow{r}-\overrightarrow{r}’|^2}\hat{r}dV$$ where (\rho(\overrightarrow{r}’) ) is the charge density at a volume element (dV) situated at (\overrightarrow{r}’ ) and (\hat{r}) is the unit vector pointing from the volume element to the observation point (P).

5. Calculate the potential energy of a charge placed in an electric field due to a continuous charge distribution.

• To calculate the potential energy of a charge (q) at a point P in an electric field due to a continuous charge distribution, we use the following expression: $$U_P=\frac{1}{4\pi\epsilon_0}q\int\frac{\rho(\overrightarrow{r}’)}{|\overrightarrow{r}-\overrightarrow{r}’|}dV$$

• Here (\rho(\overrightarrow{r}’)dV ) is the charge enclosed within the volume element (dV), (r) is the distance between (P) and the volume element (dV) , and (\epsilon_0) is the permittivity of free space.

6. Determine the electric flux through a surface due to a continuous charge distribution.

• The electric flux through a surface in the presence of a continuous charge distribution can be calculated using Gauss’s law $$ \Phi =\int \vec{E} \cdot \hat{n} dA = \frac{Q_{in}}{\epsilon_0} $$
• Here, (Q_{in}) is the total charge enclosed by the surface, (\hat{n}) is the outward normal to the surface, and (\epsilon_0) is the permittivity of free space.