Question

A point charge \(Q\) is placed a distance \(R\) from the center of a conducting sphere of radius \(a,\) with \(R>a\) (the point charge is outside the sphere). The sphere is grounded, that is, connected to a distant, unlimited source and/or sink of charge at zero potential. (Neither the distant ground nor the connection directly affects the electric field in the vicinity of the charge and sphere.) As a result, the sphere acquires a charge opposite in sign to \(Q\), and the point charge experiences an attractive force toward the sphere. a) Remarkably, the electric field outside the sphere is the same as would be produced by the point charge \(Q\) plus an imaginary mirror-image point charge \(q\), with magnitude and location that make the set of points corresponding to the surface of the sphere an equipotential of potential zero. That is, the imaginary point charge produces the same field contribution outside the sphere as the actual surface charge on the sphere. Calculate the value and location of \(q\). (Hint: By symmetry, \(q\) must lie somewhere on the axis that passes through the center of the sphere and the location of \(Q .)\) b) Calculate the force exerted on point charge \(Q\) and directed toward the sphere, in terms of the original quantities \(Q, R,\) and \(a\) c) Determine the actual nonuniform surface charge distribution on the conducting sphere.

Short answer

Question: Calculate the force exerted on point charge Q directed towards the grounded conducting sphere and determine the surface charge distribution on the conducting sphere as a function of the angle θ. Answer: The force exerted on point charge Q is given by:$$ F =\frac{Q^2}{4\pi\epsilon_0}\frac{a}{R^3}. $$The surface charge distribution on the conducting sphere is given by:$$ \sigma(\theta) = \epsilon_0 E(\theta)\cos\theta, $$where$$ E = \frac{Q}{4\pi\epsilon_0}\left(\frac{1}{(R-a\cos\theta)^2} + \frac{a}{(a\frac{a^2}{R-a} - a\cos\theta)^2}\right). $$

Step-by-step solution

Introduce image charge method and find the value and location of q

To find the equivalent point charge q, we can use the "image charge" method. By symmetry, the image charge q is located on the axis that passes through the center of the sphere and the location of Q, at a distance x from the center of the sphere. The potential at the surface of the sphere will be zero, so we can write the equation:$$ 0 = \frac{1}{4 \pi \epsilon_{0}}\left(\frac{Q}{R-x} - \frac{q}{a+x}\right). $$Now, we need to solve this equation for the value of q and x.

Solve the equation for q and x.

Considering the given equation$$ 0 = \frac{1}{4 \pi \epsilon_{0}}\left(\frac{Q}{R-x} - \frac{q}{a+x}\right),$$Prepare to solve for q and x, we get that$$ Q(R-x) = q(a+x).$$Now let's calculate q and x:$$ q = Q\frac{R-x}{a+x},$$$$ x = a\frac{R-Q/q-1}{1+Q/q}. $$Now, by rearranging the equation for q, we obtain also the expression for x:$$ q = -\frac{aQ}{R-a}. $$Plugging the value of q we found into the equation for x, we obtain:$$ x = \frac{a^{2}}{R-a}. $$

Calculate the force exerted on Q

Now that we have the value and location of the image point charge q, we can calculate the force experienced by the point charge Q. According to Coulomb's Law, the force experienced by Q is:$$ F = \frac{1}{4\pi\epsilon_0}\frac{Qq}{(R+x)^2}. $$Substituting the value of q and x we found earlier, we have:$$ F = \frac{1}{4\pi\epsilon_0}\frac{Q(-\frac{aQ}{R-a})}{(R+\frac{a^2}{R-a})^2}. $$Now, simplify the expression to find the force F:$$ F =\frac{Q^2}{4\pi\epsilon_0}\frac{a}{R^3}. $$

Determine the surface charge distribution on the conducting sphere

Since the sphere is conducting and grounded, the charge distribution on its surface will be nonuniform. To find the charge density on the surface as a function of the angle \(\theta\), we should consider that the electric field outside the sphere is the same as would be produced by point charges Q and q which has been determined as -aQ/(R-a). Thus the electric field:$$ E = \frac{Q}{4\pi\epsilon_0}\left(\frac{1}{(R-a\cos\theta)^2} + \frac{a}{(a\frac{a^2}{R-a} - a\cos\theta)^2}\right). $$The surface charge density \(\sigma(\theta)\) at any point on the sphere can be given as:$$ \sigma(\theta) = \epsilon_0 E(\theta)\cos\theta. $$By substituting the expression of the electric field E into the equation for \(\sigma(\theta)\) and simplifying, we obtain the surface charge distribution as a function of the angle \(\theta\) on the conducting sphere.

Key concepts

Conducting Sphere

A conducting sphere is a special type of material where charges can move freely across its surface. When a charge is placed near it, like the point charge in this problem, the sphere rearranges its own surface charges. This rearrangement is due to the electric field of the external charge, causing the sphere to develop an induced charge that counters the effect of the one nearby.

Because the sphere is grounded, any excess charge can flow to or from a distant location, keeping the potential at the surface at zero. This means the sphere becomes a surface of constant electric potential, or equipotential. It's important to note that the surface's final charge distribution isn't uniform, which affects how forces act between the sphere and nearby charges.

Electric Field

The electric field represents the force per unit charge exerted at any point in space by another charge. For our problem, the electric field outside the conducting sphere is influenced by both the real point charge, denoted as \(Q\), and an imaginary or 'image' charge, \(q\).

Housed entirely by the sphere's reaction to maintain zero potential on its surface, the field outside mimics that of the point charge and the image charge fixed on an imaginary line through the center. This concept simplifies calculations as it provides a way to predict how the field affects other charges outside the sphere without directly observing intricate charge distributions on the sphere's surface.

• The field arises due to the charge \(Q\).
• The image charge \(q\) ensures zero potential on the sphere's surface.

By calculating these fields, we can now predict forces and interactions in this system.

Coulomb's Law

Coulomb's Law is critical in understanding how charges interact. It defines the force between two point charges as directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them:
\[F = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r^2}\]
In our example, it helps determine the force exerted on the point charge \(Q\) due to the image charge \(q\). By substituting the known values for \(q\) and its location, we can derive the force acting toward the sphere. This approach offers a simplified yet precise method to evaluate interactions without delving into complex surface charge behaviors. Understanding Coulomb's Law ensures we appreciate how changes in distance and charge magnitude impact force strength.

Surface Charge Distribution

When an external point charge affects a grounded conducting sphere, the charges on the sphere's surface redistribute to maintain a zero potential. This distribution is usually nonuniform, reflecting how the electric field impacts different areas of the sphere.

The variable surface charge distribution is a reaction to balance the forces induced by the point charge. Using calculations based on the electric field at each point on the sphere, we derive the surface charge density \(\sigma(\theta)\):

• The external field influences how the surface charges align.
• The grounded sphere allows redistribution to maintain equilibrium.

Understanding this concept is key to predicting how the sphere's charges counterbalance any external influence.