Question

A point charge of \(+2.0 \mu C\) is located at \((2.5 \mathrm{~m}, 3.2 \mathrm{~m})\) A second point charge of \(-3.1 \mu \mathrm{C}\) is located at \((-2.1 \mathrm{~m}, 1.0 \mathrm{~m})\). a) What is the electric potential at the origin? b) Along a line passing through both point charges, at what point(s) is (are) the electric potential(s) equal to zero?

Short answer

Question: Calculate the electric potential at the origin due to two point charges, \(q_1 = +2.0 \mu C \) at (2.5m, 3.2m) and \(q_2 = -3.1 \mu C \) at (-2.1m, 1.0m). Determine the points along the line passing through both point charges where the electric potential is zero.

Step-by-step solution

Identify the given information

We are given two point charges with given locations in the plane: \(q_1\) is a charge of \(+2.0 \mu C\) at \((2.5m, 3.2m)\) and \(q_2\) is a charge of \(-3.1 \mu C\) at \((-2.1m, 1.0m)\).

Find the electric potential at the origin

We will use the principle of superposition and the electric potential formula for a point charge. The electric potential at a point due to a point charge is given by: $$V=\frac{kq}{r}$$ where \(k = 8.99\times10^9 Nm^2/C^2\) is the electrostatic constant, \(q\) is the charge, and \(r\) is the distance between the point and the charge. We find the individual potentials at the origin \((0,0)\) due to both charges and then add them together: $$V_{total} = V_1 + V_2$$ For \(q_1 = +2.0\mu C\), the distance from the origin is: \(r_1 = \sqrt{(2.5)^2 + (3.2)^2}\) For \(q_2 = -3.1\mu C\), the distance from the origin is: \(r_2 = \sqrt{(-2.1)^2 + (1.0)^2}\) Find the potential from both charges and sum them: $$V_{total} = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$$ Plug in the values and compute the result.

Determine the points where the electric potential is zero

If the total electric potential at a point along the line between the two charges is zero, then the electric potential from both charges must be equal in magnitude and opposite in sign at that point: $$\frac{kq_1}{r_1'} = - \frac{kq_2}{r_2'}$$ We can simplify this by multiplying both sides of the equation by \(r_1' r_2'\) and cancelling the electrostatic constant \(k\): $$q_1r_2'= -q_2r_1'$$ Solve for \(r_1'\) as a function of \(r_2'\): $$r_1' = \frac{-q_2}{q_1} r_2'$$ We now need to find a point(s) along the line where \(r_1'\) is proportional to \(r_2'\) in this form. We can use the equation of a line to express the positions of the point charges: $$P(x, y) = t(q_2-x_1, y_2-y_1) + q_1(x_1, y_1)$$ where \(t\) is a parameter that determines the position of the point \(P\) along the straight line between \(q_1\) and \(q_2\). Now, for any point on the path, we can compute the distances \(r_1'\) and \(r_2'\) and set their ratio to the value derived above. We can then solve for the parameter \(t\) and use it to find the position of the point(s) where the electric potential is zero. In summary, we first calculated the electric potential at the origin and then determined the points on the path through the charges where the electric potential is zero.

Key concepts

Understanding Point Charges

The concept of a point charge is fundamental to electrostatics, the branch of physics that deals with the study of stationary electric charges. A point charge is an idealized model of a charged object where the size of the object is infinitesimally small compared to the distance from the point at which the electric potential is being measured.

Imagine a point charge as if it were a tiny dot that carries an electric charge, either positive or negative. The effect this point charge has on its surroundings can be described by its electric potential, which is the work done in bringing a positive unit charge from infinity to the point under consideration.

In the exercise given, we have two point charges with specific values and locations. Taking these charges to be point charges allows us to calculate the electric potential at any other point in space, particularly at the origin, by using the formula \(V = \frac{kq}{r}\), where \(r\) is the distance between the point charge and the point where the potential is measured.

The Superposition Principle

The superposition principle is a key tenet in physics, and it states that when multiple forces or fields are acting at a single point in space, the total effect can be found by adding the individual effects linearly. In the context of electric potential, this means the total potential caused by several charges is simply the sum of the potentials due to each charge taken separately.

When applying the superposition principle to our exercise to find the electric potential at the origin, we calculate the potential due to each point charge individually and then add them up. This is because the potential is a scalar quantity, which doesn't have direction, making the summation straightforward without needing to worry about vector directions and angles.

The equation \(V_{total} = V_1 + V_2\) encapsulates this principle perfectly, representing the total electric potential at a point as the sum of potentials due to charge \(q_1\) and charge \(q_2\). This method allows us to solve for complex systems with multiple charges by breaking them down into simpler, single-charge problems.

The Electrostatic Constant

In our equations for the electric potential, \(k\) represents the electrostatic constant, also known as Coulomb's constant. This constant is crucial in determining the strength of the electric force and potential between charges. Its value is approximately \(8.99 \times 10^9 Nm^2/C^2\), and it originates from Coulomb's law, which provides the magnitude of the electrostatic force between two point charges.

The electrostatic constant offers a measure of how the electric potential will change with distance and magnitude of the point charges. The larger the charge or the smaller the distance from the charge to the point under consideration, the greater the electric potential will be.

This constant ensures the units are consistent across the formula, converting the product of charge in coulombs and distance in meters into an electric potential in volts. In the exercise and solution presented, \(k\) plays a significant role in calculating the individual potentials at the origin from each point charge.