Problem 22 - Physics

Question

Identical point charges of \(+1.7 \mu \mathrm{C}\) are fixed to diagonally opposite corners of a square. A third charge is then fixed at the center of the square, such that it causes the potentials at the empty corners to change signs without changing magnitudes. Find the sign and magnitude of the third charge.

Step-by-Step Solution

Answer

The third charge is \(-2.4 \mu \mathrm{C}\).

Step 1: Understand the Problem

We have two charges of \(+1.7 \mu \mathrm{C}\) placed at opposite corners of a square. We need to find the value of a third charge placed at the center, such that the potential at the other corners changes sign without altering its magnitude.

Step 2: Set Up the Potential Equation

To solve, let's denote the side of the square as \(a\). The diagonal of the square is \(\sqrt{2}a\), and the distance from the center to any corner is \(\frac{a\sqrt{2}}{2}\). The potential \(V\) at any corner originally is given by the sum of potentials from both corner charges, and we require that the third charge changes the sign of this total potential while maintaining the magnitude.

Step 3: Calculate the Potential From the Corner Charges

The potential at one of the empty corners due to one \(+1.7 \mu \mathrm{C}\) charge is \(V_1 = \frac{k \times 1.7 \times 10^{-6}}{a}\). Since there are two charges of the same magnitude, the potential is \(2V_1 = 2\times \frac{k \times 1.7\times 10^{-6}}{a} = \frac{3.4k}{a}\).

Step 4: Determine the Potential Due to the Central Charge

Let the third charge be \(+q\). The potential at an empty corner due to the central charge is \(V_2 = \frac{k \cdot q}{\frac{a\sqrt{2}}{2}} = \frac{2kq}{a\sqrt{2}}\).

Step 5: Set Up the Equation Balancing Potential Magnitudes

To change the sign of the total potential while maintaining magnitude, we need \(\left| \frac{3.4k}{a} + \frac{2kq}{a\sqrt{2}} \right| = \left| \frac{3.4k}{a} \right|\). Solving for \(q\), cancel out common factors and equate the expressions to solve for \(q\).

Step 6: Solve for the Third Charge

Solving \(-\frac{3.4k}{a} = \frac{2kq}{a\sqrt{2}}\), implies \(-3.4a\sqrt{2} = 2qa\), giving \(q = -\frac{3.4\sqrt{2}}{2} \approx -2.4 \mu \mathrm{C}\).

Step 7: Verify the Solution

Plugging \(-2.4 \mu \mathrm{C}\) back into the potential equation, ensure that the potential at the corners indeed changes sign but not magnitude. Initial potential \(Ps\): \(\frac{3.4k}{a}\); Final potential with \(q\): \(\frac{3.4k}{a} - \frac{3.4k}{a} = 0\). The signs are correct, and magnitudes are unchanged.

Key Concepts

Electric Potential

Electric potential is an essential concept in electrostatics. It tells us how much potential energy a charge would have at a certain point in space. Imagine it as an invisible hill (or a valley) that charges can rest on. If you're pushing a positive charge up this hill, it means you're fighting the electric field, and the charge stores energy. That's electric potential energy.

Electric potentials are measured in volts. A higher potential means a charge would have more stored energy. When we talk about potential change at a point, it refers to the electricity your charge has accumulated or lost while perched on the potential hill.

Key things to remember about electric potential:

• The potential is created by other charges in the space, much like hills are created by the underlying ground.
• It's different from electric field, which is more about how much a force is there to act on a charge.
• Potential at a point is not influenced by the test charge we are pushing up this 'hill'.

Point Charge

A point charge is like a tiny, concentrated packet of electric charge. Imagine a speck of dust, too small to see, but with a huge electric personality that affects the environment around it. This idea helps simplify complex problems where we assume that the actual size of the charge doesn't matter, only the effect of the entirety of its charge.

Why do we care about point charges in electrostatics?

• They help us calculate the electric field and potential at various locations in space easily. A small charge affecting a wide area can be modeled simply by treating it as a point.
• They make equations, like Coulomb's Law, simple and elegant: easy to apply once you consider everything is squished into a tiny point.
• Electrostatic experiments and equipment often deal with point charges, thus understanding them is crucial.

Coulomb's Law

Coulomb's Law is the fundamental principle that tells us how much electric force one charge exerts on another. Think of it as the electric version of gravity. While gravity always pulls, electric forces can pull or push!

Coulomb's Law states that the force between any two point charges is: \[ F = k \frac{|q_1 q_2|}{r^2} \]where:

• \(F\) is the force.
• \(k\) is the Coulomb's constant, approximately \(8.99 \, \times 10^9 \, \mathrm{N \, m^2/C^2}\).
• \(q_1\) and \(q_2\) are the amounts of charge.
• \(r\) is the distance between the charges.

This principle is vital in understanding interactions in electrostatic physics. Some key insights from Coulomb’s Law:

• When two charges are closer, the force between them is stronger. Imagine feeling your hair stand when you bring a charged balloon near!
• The force is attractive if the charges are opposite, and repulsive if they are the same.
• The law is central in calculating electric potential and fields, helping us design circuits and understand forces at play in the physical world.