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.

Step by step solution

01

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.

02

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:$$ x = \frac{a^{2}}{R-a}. $$

03

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:$$ F =\frac{Q^2}{4\pi\epsilon_0}\frac{a}{R^3}. $$

04

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.

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