Question

A solid conducting sphere of radius \(R_{1}\) has a charge of \(Q=4.263 \mu C\) evenly distributed over its surface. A second solid conducting sphere of radius \(R_{2}=0.6239 \mathrm{~m}\) is initially uncharged and at a distance of \(10.00 \mathrm{~m}\) from the first sphere. The two spheres are momentarily connected with a wire, which is then removed. The resulting charge on the second sphere is \(1.162 \mu C .\) What is the radius of the first sphere?

Short answer

Answer: The radius of the first sphere is approximately \(1.285 \mathrm{~m}\).

Step-by-step solution

Write the conservation of charge equation for both spheres

Since charge is conserved, the total charge before and after connecting the spheres remains the same. Let \(Q_1\) be the charge on sphere 1, and \(Q_2\) be the charge on sphere 2. Before connecting the wire, \(Q_1 = Q\) and \(Q_2 = 0\). After connecting the wire, the total charge is still \(4.263 \mu C\). Thus, we have: \(Q_1 + Q_2 = Q\) After the spheres are disconnected, we know that the charge on the second sphere is \(1.162 \mu C\). We can now write: \(Q_1 + 1.162 \mu C = 4.263 \mu C\)

Solve for the charge on the first sphere after the wire removal

Now, we will solve the equation from step 1 for \(Q_1\) to find the charge on the first sphere after the wire is removed: \(Q_1 = 4.263 \mu C - 1.162 \mu C = 3.101 \mu C\) So, the charge on the first sphere after the wire is removed is \(3.101 \mu C\).

Calculate the electric fields outside both spheres

We know that the electric fields outside a conducting sphere can be found using the following formula: \(E = \frac{kQ}{r^2}\), where \(k = 8.99 \times 10^9 \frac{N \times m^2}{C^2}\) is Coulomb's constant, \(Q\) is the charge on the sphere, and \(r\) is the distance from the center of the sphere. After the wire is removed, the electric fields outside the two spheres are equal. Let's denote them as \(E_1\) for sphere 1 and \(E_2\) for sphere 2; so, we have: \(E_1 = E_2\)

Setup the equation for the two electric fields

Now we can plug in the charge and radius expressions for each sphere. Since \(E_1 = E_2\), we get: \(\frac{kQ_1}{R_1^2} = \frac{kQ_2}{R_2^2}\) Divide both sides by \(k\) to get: \(\frac{Q_1}{R_1^2} = \frac{Q_2}{R_2^2}\)

Solve for the radius of the first sphere

Now we have the equation with one unknown, \(R_1\). Plug in the values for \(Q_1\), \(Q_2\), and \(R_2 = 0.6239 \mathrm{~m}\): \(\frac{3.101 \mu C}{R_1^2} = \frac{1.162 \mu C}{(0.6239 \mathrm{~m})^2}\) Now, solve for \(R_1\): \(R_1^2 = \frac{3.101 \mu C}{1.162 \mu C} \times (0.6239 \mathrm{~m})^2\) \(R_1^2 \approx 1.65298 \mathrm{~m^2}\) \(R_1 = \sqrt{1.65298 \mathrm{~m^2}} \approx 1.285 \mathrm{~m}\) So, the radius of the first sphere is approximately \(1.285 \mathrm{~m}\).

Key concepts

Conservation of Charge

Understanding the conservation of charge is essential when studying electricity. This principle states that the total charge in an isolated system remains constant over time, regardless of any changes that occur within the system. It's akin to the conservation of energy or mass, just applied to electrical charges. Think of it this way: if you have a certain number of charged particles, then no matter how you shuffle them around, the total charge will stay the same.

In the case of our exercise, we had two conducting spheres, one charged and one initially uncharged. When connected by a wire, electrons move from the charged sphere to the uncharged sphere until the electric potential (which depends on charge and radius) on both spheres became equal. Once the wire is removed, the rearranged charges stay on their respective spheres. Yes, the charge distribution changed, but the total charge remained the same before and after the wire connected the spheres, perfectly illustrating conservation of charge.

Electric Field

The electric field is a fascinating concept in the world of physics. It's an invisible force field that surrounds electric charges and exerts force on other electric charges within the field. The strength of this field is dependent on the size of the charge and inversely related to the square of the distance from the charge. This relationship is highlighted by the electric field equation, \( E = \frac{kQ}{r^2} \), where \( E \) represents the electric field strength, \( Q \) is the charge, \( r \) the distance from the charge, and \( k \) is Coulomb's constant.

In our textbook exercise, after detaching the connecting wire, the spheres' charge and the electric field around them had to be accounted for. Since both spheres are in their own separate space, the equation to calculate the electric field becomes pivotal. It ensures that we can describe the spaces around each sphere correctly, considering the charge they now hold.

Coulomb's Constant

Coulomb's constant, denoted by \( k \), plays a starring role in the calculations related to electric fields and forces. Its value is approximately \( 8.99 \times 10^9 \frac{N \times m^2}{C^2} \) and it's derived from Coulomb's law, which describes the force between two point charges. Understanding this constant is key to solving many problems in electrodynamics, as it relates the electric force between charges to the distance separating them.

In our solid conducting sphere problem, Coulomb's constant is a part of the formula used to equate the electric fields outside the two spheres. It serves as the bridge between the abstract concept of charge and tangible forces and electric fields we aim to calculate, ultimately guiding us toward finding the unknown radius of the first sphere.