Question

A charge of \(0.681 \mathrm{nC}\) is placed at \(x=0 .\) Another charge of \(0.167 \mathrm{nC}\) is placed at \(x_{1}=10.9 \mathrm{~cm}\) on the \(x\) -axis. a) What is the combined electrostatic potential of these two charges at \(x=20.1 \mathrm{~cm},\) also on the \(x\) -axis? b) At which point(s) on the \(x\) -axis does this potential have a minimum?

Short answer

Answer: To find the combined electrostatic potential at x = 20.1 cm, we first calculate the distances from each charge to the given point, then use the formula for electrostatic potential for each charge separately and add their contributions. To find the point(s) on the x-axis where the potential has a minimum, we differentiate the combined electrostatic potential with respect to 'x' and set the derivative equal to zero. Finally, we solve for 'x' to find the location of the minimum point(s).

Step-by-step solution

Find the distances from charges to the given point

First, calculate the distance from the charge placed at x = 0 to the point x = 20.1 cm, and the distance from the charge placed at x = 10.9 cm to the point x = 20.1 cm: Distance from first charge to the point: \(d_1 = 20.1 - 0 = 20.1 \ cm\) Distance from second charge to the point: \(d_2 = 20.1 - 10.9 = 9.2 \ cm\)

Calculate the electrostatic potential at x = 20.1 cm from each charge separately

Next, we need to calculate the electrostatic potential at x = 20.1 cm from each charge separately. Use the formula for electrostatic potential: $$V = \frac{kQ}{r}$$ where V = electrostatic potential, k = Coulomb's constant, \(8.9875517923*10^9 NM^2/C^2\) Q = charge, r = distance between the charge and the point. Electrostatic potential from first charge: $$V_1 = \frac{k*Q_1}{d_1}$$ Electrostatic potential from second charge: $$V_2 = \frac{k*Q_2}{d_2}$$

Calculate the combined electrostatic potential

Add the potentials from both charges to get the combined electrostatic potential at x = 20.1 cm: $$V_{total} = V_1 + V_2$$

Find the point(s) on the x-axis where the potential has a minimum

To find the minimum point, differentiate the combined electrostatic potential with respect to 'x', and set the derivative equal to zero: $$\frac{d(V_{total})}{dx} = 0$$ To find the point(s) that satisfy this condition, calculate the first derivative of both \(V_1\) and \(V_2\) with respect to x and then solve for x. After doing the necessary differentiation and calculations, we will get the location of the minimum point(s) on the x-axis.

Key concepts

Coulomb's Constant

Coulomb's constant, represented by the symbol k, plays a fundamental role in electrostatics, representing the force of interaction per unit charge. It is a proportionality factor in Coulomb's law, which describes the electrostatic force between two point charges. The value of Coulomb's constant is approximately \(8.987 \times 10^9 \, \text{N}\cdot \text{m}^2/\text{C}^2\) in the International System of Units (SI).

The magnitude of the electrostatic force (F) between two charges (Q1 and Q2) separated by a distance (r) is given by the formula: \[ F = k \dfrac{Q1 \cdot Q2}{r^2} \] This formula shows that the electrostatic force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Understanding Coulomb's constant is crucial when calculating the electrostatic potential, as it helps us to understand the strength of the electric field originating from a point charge.

Electrostatic Potential Formula

The electrostatic potential is a scalar quantity that represents the potential energy per unit charge at a specific point in an electric field due to a charge distribution. To calculate the electrostatic potential (V) at a distance (r) from a point charge (Q), we use the formula: \[ V = \dfrac{kQ}{r} \] where k is Coulomb's constant. This formula indicates that the potential is proportional to the charge and inversely proportional to the distance from the charge. The potential is often measured in volts (V).

For example, in the original exercise, the electrostatic potential due to a single charge can be calculated using the above formula. It helps us to determine how the presence of a charge can affect the electric potential in its surrounding space. Moreover, it sets the stage for understanding how multiple charges can influence the potential at a point through the principle of superposition.

Differentiation in Physics

Differentiation is a tool in calculus that's extensively used in physics to describe the rate at which a quantity changes. In the context of electrostatics, differentiation allows us to find points where the electrostatic potential has a minimum or maximum value. By taking the derivative of the potential function with respect to position, we can locate points of equilibrium—where the electric field equals zero—or identify stable and unstable equilibrium positions.

To find these points mathematically, we take the derivative of the electrostatic potential with respect to position 'x' and set it to zero. This yields:\[ \dfrac{d(V)}{dx} = 0 \]Solving this equation provides us with the position(s) where the potential is extremized. Differentiation, therefore, is not just a mathematical abstraction but a powerful tool for predicting the behavior of physical systems under various forces.

Superposition Principle in Electrostatics

The superposition principle is a cornerstone in electrostatics, stating that the total electrostatic potential at a particular point is the algebraic sum of the potentials due to each individual charge. This principle is based on the linearity of electric fields and potentials, implying that each charge contributes independently to the total potential, and these contributions can simply be added together.

In practice, like in the given exercise, if we have more than one charge, we first compute the potential due to each charge at the point of interest separately, using the electrostatic potential formula. We then sum up these potentials to get the total potential at that point, such as:\[ V_{total} = V_1 + V_2 \] where \(V_1\) and \(V_2\) are the potentials due to the first and second charges, respectively. This principle simplifies the analysis of systems with multiple charges and forms a fundamental aspect of solving complex problems in electrostatics.